The average temperature change per hour should be represented by the value such as = 3.2 °f /hr
How to calculate the average temperature change per hour?The temperature decrease of 20.8f° = 6.5 hrs
The decrease of temperature of xf° = 1 hr
Mathematically;
20.8°f = 6.5 hrs
X °f = 1 HR
Make X the subject of formula;
X = 20.8/6.5
= 3.2 °f /hr
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WHAT VALUE OF X WOULD MAKE THE INEQUALITY 55<6X-2 TRUE
The inequality is x > 9.5 and the numbers greater than 9.5
Given data ,
Let the inequality equation be represented as A
Now , the value of A is
55 < 6x - 2
Adding 2 on both sides , we get
6x > 57
Divide by 6 on both sides , we get
x > 9.5
So , x satisfies all numbers greater than 9.5
Hence , the inequality is x > 9.5
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The number of people who visited a winter carnival
during the first 7 hours of a day was recorded, as
shown.
79, 83, 50, 69, 86, 77, 88
It was later found that the number of people who
visited during 4th hour was incorrectly recorded. It
should have been 96. Enter a number in each box
to make the statements true.
The range of the incorrectly recorded data is
The actual range of the data is
Answer: The range of the incorrectly recorded data is:
96 - 50 = 46
The actual range of the data is:
96 - 50 = 46
The range is not affected by the correction of the 4th hour data because the range only depends on the difference between the highest and lowest values, which remains the same.
Step-by-step explanation:
Suppose that a population grows according to a logistic model with carrying capacity 5900 and k = 0. 0017 per year.
(a) Write the logistic differential equation for these data.
dP/dt =
(b) Program a calculator or computer or other tool to use Euler's method with step size h = 1 to estimate the population after 50 years if the initial population is 1000. (Round your answer to the nearest whole number. )
(c) If the initial population is 1000, write a formula for the population after years.
P(t) =
(d) Use it to find the population after 50 years. (Round your answer to one decimal place. )
(a)The logistic differential equation for these data.
dP/dt = 0.0017P(1 - P/5900)
(b) After 50 years, we would need to repeat this process 50 times to get an estimate of the population.
(c) P(t) = 6900/(1 + 5.882[tex]e^(-0.0017t))[/tex]
(d) The population after 50 years is 5869.4.
(a) The calculated differential condition for populace development is:
dP/dt = kP(1 - P/K)
where P is the populace, t is time, k is the development rate, and K is the carrying capacity.
Substituting the given values, we get:
dP/dt = 0.0017P(1 - P/5900)
(b) Utilizing Euler's strategy with step measure h = 1, we have:
P(0) = 1000
P(1) = P(0) + hdP/dt(P(0))
P(1) = 1000 + 10.00171000(1 - 1000/5900)
P(1) ≈ 1008
After 50 years, we would rehash this prepare 50 times to induce a gauge for the population.
(c) To discover an equation for the populace after a long time, able to utilize the calculated condition with starting condition P(0) = 1000. Joining both sides, we get:
∫(1/P) dP = ∫k(1 - P/K) dt
ln|P| = kt - ln|K - P|
Utilizing the starting condition, we get:
ln|1000| = k0 - ln|K - 1000|
ln|K - 1000| = ln|K| + ln|1000|
ln|K - 1000| = ln|K1000|
K - 1000 = K1000/e^(k0)
K - 1000 = K*1000/1
K = 6900
In this manner, the equation for the populace after a long time is:
P(t) = 6900/(1 + 5.882e^(-0.0017t))
(d) To discover the populace after 50 a long time, able to utilize the equation:
P(50) = 6900/(1 + 5.882e^(-0.001750))
P(50) ≈ 5869.4
The populace after 50 a long times is around 5869.4.
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the melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in a sample mean of 94.32. assume the distribution of melting point is normal with a population standard deviation of 1.20. does the true mean melting point differ from 95? use a significance level of 0.01
We do not have enough evidence to conclude that the true mean melting point differs from 95 at a significance level of 0.01.
To determine if the true mean melting point differs from 95, we can use a one-sample t-test with a significance level of 0.01.
The null hypothesis is that the true mean melting point is equal to 95, and the alternative hypothesis is that the true mean melting point is different from 95.
We can calculate the t-statistic using the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
where sample size is 16, sample mean is 94.32, hypothesized mean is 95, and sample standard deviation is the same as the population standard deviation of 1.20.
Plugging in these values, we get:
[tex]t = (94.32 - 95) / (1.20 / \sqrt{(16)} ) = -2.6667[/tex]
Using a t-distribution table with 15 degrees of freedom (n-1=16-1), and a two-tailed test at a significance level of 0.01, the critical t-value is ±2.947.
Since our calculated t-value of -2.6667 falls within the acceptance region (-2.947 < -2.6667 < 2.947), we fail to reject the null hypothesis.
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A 15-cm × 20-cm printed circuit board whose components are not allowed to come into direct contact with air for reliability reasons is to be cooled by passing cool air through a 20-cm-long channel of rectangular cross section 0. 2 cm × 14 cm drilled into the board. The heat generated by the electronic components is conducted across the thin layer of the board to the channel, where it is removed by air that enters the channel at 15°C. The heat flux at the top surface of the channel can be considered to be uniform, and heat transfer through other surfaces is negligible. The velocity of the air at the inlet of the channel does not exceed 4. 95 m/s and the surface temperature of the channel remains under 50°C. Assume that the flow is fully developed in the channel. 77777 Air 3W 15°C Air channel 0. 2 cm x 14 cm Electronic components the properties of air at a bulk mean temperature at 25C p=1. 184 kg/m k = 0. 02551 W/m°C v=1. 562x10 m/s Cy=1007 J/kg. °C Pr=0. 7296 also Nu=8. 24 Calculate the maximum total power of the electronic components that can safely be mounted on this circuit board?
The maximum total power of the electronic components that can safely be mounted on this circuit board is 4.2 W.
The maximum total power of the electronic components that can safely be mounted on the circuit board is determined by the amount of heat that can be removed from the channel by the air flow without exceeding the maximum allowable temperature of 50°C on the channel surface.
To calculate the maximum total power of the electronic components, we need to determine the heat transfer rate from the channel to the air flow
where Q is the heat transfer rate, h is the convective heat transfer coefficient, A is the surface area of the channel, and ΔT is the temperature difference between the channel surface and the air.
The convective heat transfer coefficient can be calculated using the Nusselt number correlation for flow inside a rectangular channel:
Nu = 8.24
h = 8.240.02551 /0.2 = 1.048 W/m
where L is the channel's hydraulic diameter, equal to [tex]2*(0.2*14)/(0.2+14)[/tex] = 0.278 cm = 0.00278 m.
The surface area of the channel is A = 20.220 + 20.214 + 14[tex]*20[/tex] = 120.8 cm[tex]^2[/tex] = 0.01208 [tex]m^2.[/tex]
The temperature difference between the channel surface and the air is ΔT = 50°C - 15°C = 35°C.
Therefore, the maximum heat transfer rate from the channel to the air flow is:
Q = hAΔT = 1.0480.0120835 = 0.0042 kW
This means that the maximum total power of the electronic components that can be safely mounted on the circuit board is:
P = Q = 0.0042 kW = 4.2 W
Therefore, the maximum total power of the electronic components that can safely be mounted on this circuit board is 4.2 W.
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QUESTION 10 During oxygen consumption measurement the participants VO2 was 1004 L/min and VCO2 was 0.932 min. What was the participants RER at that point in time? Give your answer to 4 decimal places
The participant's RER at that point in time was approximately 0.0009 (rounded to 4 decimal places).
To calculate the participant's Respiratory Exchange Ratio (RER) at that point in time, you will use the formula:
RER = VCO2 / VO2
Given the participant's VO2 was 1004 L/min and VCO2 was 0.932 L/min, plug these values into the formula:
RER = 0.932 / 1004
Now, divide 0.932 by 1004:
RER ≈ 0.0009
So, the participant's RER at that point in time was approximately 0.0009 (rounded to 4 decimal places).
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Compare the functions shown below: Which function has the greatest y-intercept?
f(x)
b
g(x)
c
h(x)
d
All three functions have the same y-intercept.
The circle centered at Q is a scaled copy of the circle centered at R.
a. Find the scale factor.
ingrese su respuesta.....
840
The scale factor of the dilation of the circles is 5
What is the scale factor of the dilation?From the question, we have the following parameters that can be used in our computation:
The circles
From the circles, we have the following parameters
Diameter of big circle Q = 20
Diameter of the small circle R = 4
Using the above as a guide, we have the following:
Scale factor of the dilation = Radius of big circle/Radius o the small circle
So, we have
Scale factor of the dilation = 20/4
Evaluate
Scale factor of the dilation = 5
Hence, the scale factor of the dilation is 5
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Estimate the perimeter and the area of the shaded figure to the nearest tenth.
A shaded composite figure is shown on a grid. The figure is made up of a triangle on the left, a square in the center, and a semicircle on the right.The triangle is a 45, 45, 90 triangle with a hypotenuse of 6 units. The semicircle has a diameter of 6 units. The square has a side length of 2 units. One side of the square is shared with part of the hypotenuse of the triangle, and one side is shared with part of the diameter of the semicircle.
perimeter: about
units
area: about
square units
Please answer
for perimeter
and area
a) The perimeter of the shaded figure is P = 29.9 units
b) The area of the shaded figure is A = 27.14 units²
Given data ,
Let the circumference of the semicircle C = πr
On simplifying , we get
C = π ( 3 ) = 9.42 units
Now , the total number of straight lines is n = 6 with 2 units
So , perimeter of straight lines = 12 units
Each diagonal represents a hypotenuse of an equal-sided triangle with side = 3 units
So the length of each diagonal is 3√2 units
The perimeter of shaded figure P = 9.42 units + 12 units + 2 ( 3√2 units )
P = 29.9 units
b)
The area of the shaded figure is A
Now , the value of A is
A = area of semicircle + area of square + area of triangle
A = ( πr² )/2 + ( 2 )² + 2 ( 1/2 )bh
A = 14.14 + 4 + 9
A = 27.14 units²
Hence , the perimeter and area is solved
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The complete question is attached below :
Estimate the perimeter and the area of the shaded figure to the nearest tenth.
A shaded composite figure is shown on a grid. The figure is made up of a triangle on the left, a square in the center, and a semicircle on the right.The triangle is a 45, 45, 90 triangle with a hypotenuse of 6 units. The semicircle has a diameter of 6 units. The square has a side length of 2 units. One side of the square is shared with part of the hypotenuse of the triangle, and one side is shared with part of the diameter of the semicircle.
perimeter: about
units
area: about
square units
draw the graph of polynomial function
The graph of the polynomial function that passes through the points P(−2,2) and Q(1,0) is added as an attachment
Drawing the graph of polynomial functionFrom the question, we have the following parameters that can be used in our computation:
The graph passes through the points P(−2,2) and Q(1,0)
Assuming the graph is a linear function
So, we have
y = mx + c
Where
c = y when x = 0 and m = slope
This gives
-2m + c = 2
m + c = 0
Subtract the equations
So, we have
-3m = 2
m = -2/3
Solving for c, we have
-2/3 + c = 0
Evaluate
c = 2/3
So, the equation is f(x) = -2/3x + 2/3
See attachment for the graph
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Complete question
Draw the graph of polynomial function f(x) containing the points P(−2,2) and Q(1,0)
the number of square feet per house are normally distributed with a population standard deviation of 137 square feet and an unknown population mean. a random sample of 19 houses is taken and results in a sample mean of 1350 square feet. find the margin of error for a 80% confidence interval for the population mean. z0.10z0.10 z0.05z0.05 z0.025 z 0.025 z0.01z0.01 z0.005 z 0.005 1.282 1.645 1.960 2.326 2.576 you may use a calculator or the common z values above. round the final answer to two decimal places.
The margin of error for a 80% confidence interval for the population mean is 57.82 square feet.
The number of square feet per house are normally distributed with a population standard deviation of 137 square feet and an unknown population mean. To find the margin of error for a 80% confidence interval, we need to use the formula:
Margin of error = z*(σ/√n)
where z is the z-score corresponding to the level of confidence (80% corresponds to z=1.282), σ is the population standard deviation (given as 137), and n is the sample size (given as 19).
Plugging in the values, we get:
Margin of error = 1.282*(137/√19) = 57.82
Therefore, the margin of error for a 80% confidence interval is 57.82 square feet.
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Please Answer fast Enhancer 1 Find the temperature of the sun if pressure is 1.4x10 atm, density is 1.4 g/cc and average molecular weight of gases present there is 2 (R = 8.4 x 107 erg/mol/K) (a) 3.2 x 10'K (b) 2.4 x 10'K (c)1.2 x 10K (d) 1.8 x 107K
To find the temperature of the sun, we'll use the ideal gas law equation, which is PV = nRT. We're given pressure (P), density (ρ), average molecular weight (M), and the gas constant (R). First, we'll find the number of moles (n) and then solve for temperature (T). After the calulation the answer is found out to be option b which is approximately 2.4 x 10^7 K.
1. Calculate the number of moles (n) using the formula n = ρ/M.
n = 1.4 g/cc / 2 g/mol = 0.7 mol/cc
2. Rearrange the ideal gas law equation to solve for temperature (T): T = PV / nR
3. Plug in the values:
P = 1.4 x 10^10 atm
V = 1 cc (since we are considering 1 cc of the gas)
n = 0.7 mol
R = 8.4 x 10^7 erg/mol/K
T = (1.4 x 10^10 atm) x (1 cc) / (0.7 mol) x (8.4 x 10^7 erg/mol/K)
4. Perform the calculation:
T = 1.4 x 10^10 / (0.7 x 8.4 x 10^7) = 2.38 x 10^7 K
The temperature of the sun is approximately 2.4 x 10^7 K (option b).
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Find the mean, median, mode, range, and standard deviation of the data set that is obtained after multiplying each value by the given constant. Round to the nearest tenth, if necessary. If there is no mode, write none.
1, 5, 4, 2, 1, 3, 6, 2, 5, 1; x6.5
Answer:
Step-by-step explanation:
First, organize the numbers; 1,1,1,2,2,3,4,5,5,6. The median is the middle number, in this case, 2 and 3 are in the middle. You'd add 2+3 which equals 5, then divide by 2 to get 2.5. The median is 2.5, to find the mode you need to see how many times one number appears. The mode is 1, the range is the largest number minus the smallest number. The range is 5, to find the mean you add all numbers up, then divide by how many numbers there are. All the numbers added up are 30, now divide that by 10 to get 3. The mean is 3. The standard deviation is 1.7 x 6.5 = 20.8
I'm only an algebra student so I may not be entirely correct.
An orange cone has a volume of 376.8 cubin cm and a radius of 6 cm. What is the height?
Answer:
3.29 cm.
Step-by-step explanation:
The formula for the volume of a cone is V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
Substituting the given values, we get:
376.8 = (1/3)π(6^2)h
Simplifying:
376.8 = 36πh
h = 376.8 / (36π)
h ≈ 3.29 cm
Therefore, the height of the orange cone is approximately 3.29 cm.
One gallon of water weighs 8. 34 pounds. How much does 6 gallons of water weigh? Show your work, including labeling your answer with the appropriate units of measure
According to the given situation, 6 gallons of water weighs 50.04 pounds.
The United States and certain other nations frequently use the gallon as a unit of volume measurement. Although there are other types of gallons, the U.S. gallon, which is equivalent to 128 fluid ounces or 3.785 liters, is the most often used. It is used to calculate the volume of fluids like milk, water, petrol, and other substances. The word gallon is shortened to "gal."
One gallon of water weighs 8.34 pounds. To find the weight of 6 gallons of water, we can multiply the weight of one gallon by 6:
Weight of 6 gallons of water = 6 x 8.34 pounds
On simplifying we get:
Weight of 6 gallons of water = 50.04 pounds
Therefore, 6 gallons of water weighs 50.04 pounds.
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A class of students who have been studying relations has proposed
(incorrectly) that each of the following relations R on set A is a
partial order. For each relation, determine which property or properties
(reflexive, anti-symmetric, transitive) the relation fails to satisfy.
Sure, I can help with that! To determine which property or properties each relation fails to satisfy, we first need to understand what each of those properties means.
A relation R on a set A is reflexive if for every element a in A, (a,a) is in R.
A relation R on a set A is anti-symmetric if for every distinct elements a and b in A, if (a,b) is in R then (b,a) is not in R.
A relation R on a set A is transitive if for every elements a, b, and c in A, if (a,b) is in R and (b,c) is in R then (a,c) is in R.
Now, let's look at each of the proposed relations and determine which properties they fail to satisfy:
1. R = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}
This relation is not anti-symmetric because (1,2) is in R and (2,1) is also in R.
2. R = {(1,1), (2,2), (3,3), (1,2), (2,1)}
This relation is not transitive because (1,2) is in R and (2,1) is also in R, but (1,1) is not in R.
3. R = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (3,2)}
This relation is not anti-symmetric because (3,2) is in R and (2,3) is also in R.
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I have 25 minQuestion 9 A man swimming in a stream which flows 4 Km/h finds that in a given time he can swim 5 times as far with the stream as he can against it. At what rate does he swim? Not yet answered Marked
If a man swimming in a stream which flows 4 Km/h finds that in a given time he can swim 5 times as far at 12 km/h rate he can swim.
Production of upstream oil and gas is carried out by businesses that locate, mine, or create raw resources. The end-user or customer is closer to the production of oil and gas in the downstream sector. Here is a look at the upstream and downstream production of oil and gas, their individual roles, and how they fit into the larger supply chain.
Let's take swimmer speed = x km/h.
The time is taken by the swimmer = t
The speed of the stream = 4 km/h
The speed when swimming with the stream = x+ 4 km/h
The distance covered when swimming with the stream D₁= (x+4)t
The speed when swimming against the stream = x -4
The distance covered when swimming against the stream D₂= (x-4)t
The swimmer swims 5 times when swimming against the stream
Therefore,
D₁ = 5D₂
(x+4)t = 2((x-4)t)
x+4 = 2x- 8
x = 12 km/h
12 km/h is the speed of the swimmer.
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7. Use the following figure to answer questions a-d.
a. Find the perimeter of the figure if the following is true:
a = x + 7
b = 2x + 2
C = 8x
d = x + 10
b. What is the perimeter of the figure in part
(a) if x = 4
c. What is the perimeter of the figure in part (b)
if x = 2 ?
d. Find the perimeter of the figure if the following
is true:
a = x2
b = 4x + 8
c = 2x²
d = x
a) The perimeter of the figure [tex]p = 12x + 19[/tex]
b) The perimeter of the figure when [tex]x = 4[/tex] is 67 units.
c) The perimeter of the figure when[tex]x = 2[/tex] is 43 units
d) The perimeter of the figure is [tex]3x^2 + 5x + 8[/tex] units.
a) The perimeter P of the figure is the sum of the lengths of its sides:
[tex]P = a + b + c + d[/tex]
Substituting the given expressions for a, b, c, and d in terms of x, we get:
[tex]P = (x + 7) + (2x + 2) + (8x) + (x + 10)[/tex]
[tex]= 12x + 19[/tex]
b) Substituting x = 4, we get:
[tex]P = 12x + 19[/tex]
[tex]= 12(4) + 19[/tex]
[tex]= 67[/tex]
Therefore, the perimeter of the figure when [tex]x = 4[/tex] is 67 units.
c) Substituting [tex]x = 2,[/tex] we get:
[tex]P = 12x + 19[/tex]
[tex]= 12(2) + 19[/tex]
[tex]= 43[/tex]
Therefore, the perimeter of the figure when[tex]x = 2[/tex] is 43 units.
d) Substituting the given expressions for a, b, c, and d in terms of x, we get:
[tex]P = x^2 + (4x + 8) + 2x^2 + x[/tex]
[tex]= 3x^2 + 5x + 8[/tex]
Therefore, the perimeter of the figure when[tex]a = x^2, b = 4x + 8, c = 2x^2,[/tex] and [tex]d = x[/tex] is [tex]3x^2 + 5x + 8[/tex] units.
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In a class of 100 students, 55 students have passed in physics and 67 students have passed in Mathematics. Find the number of students passed in Physics only.
Answer:
Step-by-step by steo
Total number of students = n(P∪M)=100
Number of students who passed in physics= n(P)=55
Number of students who passed in maths= n(M)=67
Number of students who passed both physics and maths=n(P∩M)
⇒n(P∪M)=n(P)+n(M)−n(P∩M)
⇒100=55+67−n(P∩M)
⇒n(P∩M)=122−100
⇒n(P∩M)=22
Step - 2 : Find number of students who passed only in physics
Number of students passed in physics = 55
Number of students passed in physics and maths both = 22
∴Number of students passed only in physics = 55−22=33
Nachelle earned a score of 725 on Exam A that had a mean of 700 and a standard deviation of 50. She is about to take Exam B that has a mean of 100 and a standard deviation of 20. How well must Nachelle score on Exam B in order to do equivalently well as she did on Exam A? Assume that scores on each exam are normally distributed.
Nachelle needs to score 110 on Exam B in order to do equivalently well as she did on Exam A.
Given data: Nachelle earned a score of 725 on Exam A that had a mean of 700 and a standard deviation of 50.
She is about to take Exam B that has a mean of 100 and a standard deviation of 20.Let x be the score on Exam B that Nachelle needs to do equivalently well as she did on Exam A.
According to the Z-score formula, Z = (x - μ) / σ where Z is the standard score, x is the value of the element, μ is the population mean, and σ is the standard deviation.
Let's calculate the Z-scores for Nachelle's scores on Exams A and B.
Z-score for Nachelle's score on Exam AZ1 = (725 - 700) / 50 = 0.5
Z-score for Nachelle's score on Exam BZ2 = (x - 100) / 20 = (x - 100) / 20
Now, if Nachelle has to do equivalently well on Exam B as she did on Exam A, then the Z-scores for both exams should be equal.
Hence,0.5 = (x - 100) / 20Solving for x,x - 100 = 0.5 × 20 = 10x = 100 + 10 = 110.
Therefore, Nachelle needs to score 110 on Exam B in order to do equivalently well as she did on Exam A.
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If the diameter of a circle is 8.4 in., find the area and the circumference of the circle. Use 3.14 for pi. Round your answers to the nearest hundredth.
Answer:
Area of the circle = 55.39 in²
Circumference of the circle = 26.38 in
Step-by-step explanation:
Given, the diameter of the circle = 8.4
so the radius is given by the formula
∴ d = 2r
→ 8.4 = 2×r
→r=4.2 in [i]
Area of the circle = π×r² [ii]
substituting the value of r in equation [ii]
we get,
area of circle = 3.14×(4.2)²
=55.39 in²
circumference of the circle =2πr [iii]
substituting the value of r in equation [iii]
we get,
circumference of the circle = 2×3.14×4.2
= 26.38 in
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Jeff's work to find the missing length of the triangle is shown. Explain Jeff's error.
The missing length of the triangle is given as follows:
x = 35 units.
What is the Pythagorean Theorem?The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.
The theorem is expressed as follows:
c² = a² + b².
In which:
c is the length of the hypotenuse.a and b are the lengths of the other two sides (the legs) of the right-angled triangle.For the triangle in this problem, we have that:
There is a side of 12 units.The hypotenuse has length of 37 units.Hence the missing length is obtained as follows:
x² + 12² = 37²
x = sqrt(37² - 12²)
x = 35 units.
Missing InformationThe image is blurry, hence the mistake in Jeff's part cannot be explained.
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(PLEASE HELP) A student is building a squirrel feeder for a family member. The figure is a model of the feeder.
A rectangular prism with dimensions 4 and one-fourths inches by 18 and one-fourth inches by 3 inches.
How much feed can the container hold?
eighty and one-half in3
one hundred sixteen and one-sixteenth in3
two hundred thirty-two and eleven-sixteenths in3
four hundred sixty-five and three-eighths in3
The amount of feed the container can hold is 232 11/16 cubic inches. The correct option is the third option - two hundred thirty-two and eleven-sixteenths in3
Calculating how much feed the container can holdFrom the question, we are to calculate how much feed the container can hold.
From the given information, the container is a rectangular prism
To calculate how much fed the container can hold, we will determine the volume of the rectangular prism
Volume of a rectangular prism is given by the formula
Volume = Length × Width × Height
Thus,
Volume of the container = 4 1/4 × 18 1/4 × 3
Volume of the container = 232 11/16 cubic inches
Hence,
The quantity of feed it can hold is 232 11/16 cubic inches
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suppose that we repeatedly draw a random card from a standard deck of 52 cards with replacement until we draw a heart or a face card. note: in each suit, only jacks, queens, and kings are face cards. (a) what is the probability that we draw a total of 4 cards? (b) what is the probability that we draw total of n cards, where n is a positive integer? (c) what is the expected number of times we draw a card?
The probability that we draw a total of 4 cards is approximately 0.074.
The probability that we draw a total of n cards is a function of n.
the expected number of times we draw a card until we get a heart or a face card is: E(X) = 1/p = 3.25.
How we get the probability?To solve this problem, we can use the geometric distribution, which models the number of trials needed to get the first success in a sequence of independent trials.
Find the probability of success p.Since we are drawing a heart or a face card, there are 16 cards (4 face cards and 12 hearts) out of 52 that are successful. Therefore, the probability of success is:
p = 16/52 = 4/13
Use the geometric distribution to answer the questions.What is the probability that we draw a total of 4 cards
We want to find the probability that we get the first success on the fourth trial, i.e., we draw three non-successes followed by a success. The probability of this event is:
P(X = 4) = [tex](1 - p)^3 * p = (9/13)^3 * (4/13) = 0.074[/tex]
What is the probability that we draw a total of n cards, where n is a positive integerWe want to find the probability that we get the first success on the nth trial, i.e., we draw n-1 non-successes followed by a success. The probability of this event is:
[tex]P(X = n) = (1 - p)^(^n^-^1^) * p[/tex]
What is the expected number of times we draw a cardThe expected value of a geometric distribution is 1/p. Therefore, the expected number of times we draw a card until we get a heart or a face card is:
E(X) = 1/p = 13/4
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a) what is the angle of elevation from
Row A to the bottom of the screen?
b) what is the angle of depression from
Row P to the bottom of the screen?
Give your answers to 1 d.p.
Screen
2.5 m
5.6 m
12°
Row A
19.6 m
Row P
The angle of elevation from Row A to the bottom of the screen 13.3.
The angle of depression from Row P to the bottom of the screen 4.4
let angle of elevation of Row A to the bottom of the Screen be Ae
So, tan Ae = 2.5 - 5.8tan 11 /5.8
tan Ae = 0.23665
Ae = 13.3
Now, let the angle of depression of Row P to the bottom of the screen be P
tan P = 2.3 tan 11- 2.5/ 2.5
tan P = 0.077
P = 4.4
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please explain and show all steps
а Assume X is normally distributed with a mean of 11 and a standard deviation of 3 Determine the value of x that solves P (X> x) = 0.8
We can conclude that if X is normally distributed with a mean of 11 and a standard deviation of 3, then the value of x that solves P(X > x) = 0.8 is x = 13.52.
We need to find the value of x such that P(X > x) = 0.8, where X is a normally distributed random variable with mean μ = 11 and standard deviation σ = 3.
From the properties of the standard normal distribution, we know that if Z is a standard normal random variable, then P(Z > z) = 0.8 corresponds to z = 0.84 (found using a standard normal table or calculator).
We can standardize X to a standard normal random variable Z using the formula:
Z = (X - μ) / σ
Substituting the values μ = 11 and σ = 3, we get:
Z = (X - 11) / 3
Now, we want to find the value of x such that P(X > x) = 0.8. We can rewrite this as:
P(Z > (x - 11) / 3) = 0.8
Using the standard normal table or calculator, we find that P(Z > 0.84) = 0.2005.
Therefore, we can write:
0.2005 = P(Z > 0.84) = P((X - 11) / 3 > 0.84) = P(X > 11 + 3(0.84)) = P(X > 13.52)
So the value of x that solves P(X > x) = 0.8 is x = 13.52.
Therefore, we can conclude that if X is normally distributed with a mean of 11 and a standard deviation of 3, then the value of x that solves P(X > x) = 0.8 is x = 13.52.
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19. A company known for making wood bats for Major League Baseball designs the bats to last between 48 days and 72 days. The life expectancy of wood bats is normally distributed with a mean of 60 days and a
standard deviation of 5 days.
(a) What is the probability that a randomly chosen bat will last more than 70 days?
(b) What percentage of bats fail to last the designed amount of days? (48-72)
To discover the likelihood that a haphazardly chosen bat will final more than 70 days, we have to be standardize the esteem utilizing the standard typical conveyance. Ready to do this by calculating the z-score:
z = (70 - 60) / 5 = 2
Employing a standard typical dissemination table or calculator, we discover that the likelihood of a z-score more noteworthy than 2 is roughly 0.0228. Hence, the likelihood that a haphazardly chosen bat will final more than 70 days is around 0.0228.
What percentage of bats fail to last the designed amount of days?To discover the rate of bats that fall flat to final the outlined sum of days (48-72), we have to be discover the region beneath the typical distribution curve to the cleared out of 48 and to the proper of 72 and include them together. This speaks to the likelihood of a bat enduring less than 48 days or more than 72 days.
To standardize the values of 48 and 72, we utilize the same equation as in portion (a):
z1 = (48 - 60) / 5 = -2.4
z2 = (72 - 60) / 5 = 2.4
Employing a standard ordinary conveyance table or calculator, we discover that the range to the cleared out of z1 is around 0.0082 and the region to the correct of z2 is additionally around 0.0082. Hence, the full likelihood of a bat falling flat to final between 48 and 72 days is roughly:
0.0082 + 0.0082 = 0.0164
To change over this to a rate, we duplicate by 100:
0.0164 * 100 = 1.64%
Hence, roughly 1.64% of bats come up short to final the outlined sum of days.
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12x to the power of 2 y divided by 3x to the power of 2.
please help!
The value of the expression is 4y.
Given is an expression, [tex]12x^{2} y/ 3x^2[/tex], we need to simplify it,
[tex]12x^{2} y/ 3x^2[/tex]
= 12 × x² × y / 3 × x²
= 4 × x² × y / x²
= 4y
Hence, the value of the expression is 4y.
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Find the diameter of circle O
The diameter of the circle is determined as 20.12 units.
What is the radius of the circle?The radius of the circle is calculated by applying the following formula as shown below;
The let the distance between 9 and center of the circle O = x
The radius of the circle, r = 9 + x --- (1)
Draw a line from circumference from point 10 to center O, this line is the radius = r
Apply Pythagoras theorem;
r² = 10² + x² ------ (2)
From the first equation, recall, r = 9 + x
(9 + x)² = 10² + x²
81 + 18x + x² = 100 + x²
81 + 18x = 100
18x = 100 - 81
18x = 19
x = 19/18
x = 1.06
The radius of the circle = 9 + x
r = 9 + 1.06
r = 10.06
The diameter of the circle = 2r
= 2 x 10.06
= 20.12 units
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Consider a volleyball net that consists of a mesh with m squares on the horizontal dimension
and n squares on the vertical. What is the maximum number of strings that can be cut before
the net falls apart into two pieces? Solve the problem as a network problem.
The maximum number of strings that can be cut before the volleyball net falls apart into two pieces, considering a mesh with m squares on the horizontal dimension and n squares on the vertical dimension, is min(m+1, n+1).
To determine the maximum number of strings that can be cut before the volleyball net falls apart into two pieces, considering the mesh has m squares on the horizontal dimension and n squares on the vertical dimension, we can solve this as a network problem. Here's a step-by-step explanation:
1. Visualize the volleyball net as a graph with nodes representing the intersections of the strings and edges representing the strings themselves.
2. Observe that there are (m+1) nodes horizontally and (n+1) nodes vertically. This is because each square has a node at each corner, so there is an additional node on each side.
3. To separate the net into two pieces, you need to cut all the strings along a specific row or column of the mesh.
4. The maximum number of strings you can cut will be the minimum number of strings needed to cut through the entire mesh either horizontally or vertically.
5. If you choose to cut the strings horizontally, there will be (m+1) strings to cut for each row. Similarly, if you choose to cut the strings vertically, there will be (n+1) strings to cut for each column.
6. Compare the number of strings to cut in both directions, and choose the direction with the minimum number of strings to cut.
7. The maximum number of strings that can be cut before the net falls apart into two pieces is the minimum between (m+1) and (n+1).
So, the maximum number of strings that can be cut before the volleyball net falls apart into two pieces, considering a mesh with m squares on the horizontal dimension and n squares on the vertical dimension, is min(m+1, n+1).
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