For equation A+C=B the matrix C is [tex]\left[\begin{array}{ccc}-2&-7\\-5&8\end{array}\right][/tex] and C-B=A then C is [tex]\left[\begin{array}{ccc}2&-9\\7&0\end{array}\right][/tex]
The given matrix A = [tex]\left[\begin{array}{ccc}2&-1\\6&-4\end{array}\right][/tex]
B=[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]
Now the equation is A+C=B
[tex]\left[\begin{array}{ccc}2&-1\\6&4\end{array}\right][/tex]+C =[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]
C=[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]- [tex]\left[\begin{array}{ccc}2&-1\\6&-4\end{array}\right][/tex]
C=[tex]\left[\begin{array}{ccc}-2&-7\\-5&8\end{array}\right][/tex]
Now equation is C-B=A
C=A+B
= [tex]\left[\begin{array}{ccc}2&-1\\6&-4\end{array}\right][/tex]+[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]
C=[tex]\left[\begin{array}{ccc}2&-9\\7&0\end{array}\right][/tex]
Hence, for equation A+C=B the matrix C is [tex]\left[\begin{array}{ccc}-2&-7\\-5&8\end{array}\right][/tex] and C-B=A then C is [tex]\left[\begin{array}{ccc}2&-9\\7&0\end{array}\right][/tex]
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At West High School, 10% of the students participate in sports. A student wants to simulate the act of randomly
selecting 20 students and counting the number of students in the sample who participate in sports. What is an
appropriate assignment of digits for this simulation?
O Let 0-8 = the student participates in sports. Let 9 = the student does not participate in sports.
Let 0 = the student participates in sports. Let 1-9 = the student does not participate in sports.
Let 0 and 1 = the student participates in sports. Let 2-9= the student does not participate in sports.
O Let 2-9 = the student participates in sports. Let 0 and 1 = the student does not participate in sports.
Let 0-1 represent students who participate in sports and let 2-9 represent students who do not participate in sports.
Let 0-9 represent students selected for the sample, with digits 0-8 representing students who participate in sports and digit 9 representing a student who does not participate in sports.
So, an appropriate assignment of digits for this simulation would be: Let 0-1 represent students who participate in sports and let 2-9 represent students who do not participate in sports.
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Use the table to identify values of p and q that can be used to factor
x2-4x-12
as (x + p)(x+q).
OA. -2 and 6
OB. 2 and -6
OC. 3 and -4
OD. -3 and 4
P
2-6 -4
qp+q
-2 6 4
-1
1
3-4
-3 4
Answer: Use the table to identify values of p and q that can be used to factor x2 + x – 12 as (x + p)(x + q).A. –2 and 6. B. 3 and –4. C. –3 and 4. D. 2 and –6.
Step-by-step explanation:
Question 2 (5 points)
ABC is a right triangle
AC=12
CB=9
Blank #1 Find AB
Do not label
Blank #2. Find /A
Round your answer to the nearest whole number. Do not include a degree sign
Blank #3 Find /C
Round your answer to the nearest whole number. Do not include a degree sign.
Blank #4 Find /B
Round your answer to the nearest whole number. Do not include a degree sign
Question 2 options:
The required solution for the given right angle triangle is given below.
Blank #1: We can use the Pythagorean theorem to find AB:
[tex]AB = \sqrt{AC^2 + CB^2} \\= \sqrt{12^2 + 9^2}\\ =15[/tex]
Therefore, AB = 15.
Blank #2: We can use the inverse tangent function to find the angle A:
tan(A) = opposite / adjacent = CB / AC = 9 / 12
[tex]A = tan^{-1}(9/12) = 36.86^o[/tex]
Therefore, angle A ≈ 36 degrees.
Blank #3: We can use the inverse cosine function to find the angle C:
cos(C) = adjacent / hypotenuse = CB / AB = 9 / 15
C = arccos(9/15) ≈ 53.14
Therefore, angle C ≈ 53.14 degrees.
Blank #4: We can use the fact that the sum of angles in a triangle is 180 degrees to find angle B:
B = 180 - A - C ≈
B= 90
Therefore, angle B ≈ 90
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a scientist was interested in studying if students beliefs about illegal drug use changes as they go through college. the scientist randomly selected 104 students and asked them before they entered college if they thought that illegal drug use was wrong or o.k. four years later, the same 104 students were asked if thought that illegal drug use was wrong or o.k. the scientist decided to perform mcnemar's test. the data is below. what is the null hypothesis?
In this case, the scientist is studying if students' beliefs about illegal drug use change as they go through college. The null hypothesis (H0) for McNemar's test in this context would state that there is no significant change in students' beliefs about illegal drug use between the time they enter college and four years later. In other words, the proportion of students who change their beliefs about illegal drug use is not significantly different from the proportion who do not change their beliefs.
Null hypothesis (H0): There is no significant difference in students' beliefs about illegal drug use before and after going through college.
Prove or disprove. show your work.
(a) for any integers n a and m: if both n and m are odd, then n - m² is even
(b) Vp Z: if p is prime, then p-2 is not prime.
(c) Vs R s is irrational s2 is irrational.
(d) There is two odd integers n and m such that n² m² - 1 is odd.
(a) The statement is false because we have found a case where n - m² is even.
(b) The statement holds true.
(c) The statement is false because we have found a case where s^2 is rational despite s being irrational.
(d) It is not possible to find two odd integers n and m such that n²m² - 1 is odd. Thus, the statement is false.
(a) The statement "for any integers n and m, if both n and m are odd, then n - m² is even" is incorrect. Let's consider a counterexample:
Take n = 3 and m = 1. Both n and m are odd.
n - m² = 3 - 1² = 3 - 1 = 2, which is an even number.
Therefore, the statement is false because we have found a case where n - m² is even.
(b) The statement "for any prime number p, p-2 is not prime" is generally true. Let's consider the cases:
If p is an odd prime greater than 2, then p-2 is an even number, and the only even prime number is 2. Therefore, p-2 cannot be prime in this case.
If p = 2, then p-2 = 0, which is not considered a prime number.
In both cases, p-2 is not a prime number. Therefore, the statement holds true.
(c) The statement "for any real number s, if s is irrational, then s^2 is irrational" is incorrect. Let's consider a counterexample:
Take s = √2. √2 is an irrational number.
s^2 = (√2)^2 = 2, which is a rational number.
Therefore, the statement is false because we have found a case where s^2 is rational despite s being irrational.
(d) The statement "There are two odd integers n and m such that n²m² - 1 is odd" is true. Let's consider the following example:
Take n = 1 and m = 1. Both n and m are odd.
n²m² - 1 = 1² * 1² - 1 = 1 * 1 - 1 = 0, which is an even number.
However, if we take n = 3 and m = 1, both n and m are still odd.
n²m² - 1 = 3² * 1² - 1 = 9 * 1 - 1 = 9 - 1 = 8, which is an even number.
Therefore, it is not possible to find two odd integers n and m such that n²m² - 1 is odd. Thus, the statement is false.
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A circular spinner has a radius of 6 inches. The spinner is divided into three sections of unequal area. The sector labeled "green" has a central angle of 60°. A point on the spinner is randomly selected.
What is the probability that the randomly selected point falls in the green sector?
Responses
1 over 60
1 over 6
1 over 4
1 over 3
The probability that the randomly selected point falls in the green sector is 1/6.
Option B is the correct answer.
We have,
The area of the green sector can be found by using the formula for the area of a sector:
A = (θ/360)πr²,
Where θ is the central angle and r is the radius.
In this case,
θ = 60° and r = 6 inches,
So the area of the green sector is:
A = (60/360)π(6)²
A = π(6)²/6
A = 6π
So,
The total area of the spinner is π(6)² = 36π.
So the probability of the randomly selected point falling in the green sector is:
P = (Area of green sector)/(Total area of spinner)
P = (6π)/(36π)
P = 1/6
Therefore,
The probability that the randomly selected point falls in the green sector is 1/6.
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WILL GIVE BRAINLIEST if helped
the most important part of this is the first post thing
The segment length and the conversion of radian and degree are given below.
We have,
In order to solve for segment length in relation to circles, chords, secants, and tangents, we need to first define some terms:
Circle: A set of all points in a plane that are equidistant from a given point called the center of the circle.
Chord: A line segment joining two points on a circle.
Secant: A line that intersects a circle in two points.
Tangent: A line intersecting a circle at exactly one point, called the point of tangency.
Segment: A part of a circle bounded by a chord, a secant, or a tangent and the arc of the circle that lies between them.
Now, let's consider the following cases:
Chord-chord intersection:
If two chords intersect inside a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. That is:
AB × BC = DE × EF
where AB and BC are the lengths of the segments of one chord, and DE and EF are the lengths of the segments of the other chord.
Secant-secant intersection:
If two secants intersect outside a circle, the product of the length of one secant and its external segment is equal to the product of the length of the other secant and its external segment. That is:
AB × AC = DE × DF
where AB and AC are the length of one secant and its external segment, and DE and DF are the length of the other secant and its external segment.
Secant-tangent intersection:
If a secant and a tangent intersect outside a circle, the product of the length of the secant and its external segment is equal to the square of the length of the tangent. That is:
AB × AC = AD^2
where AB and AC are the length of the secant and its external segment, and AD is the length of the tangent.
Tangent-tangent intersection:
If two tangents intersect outside a circle, the lengths of the two segments of one tangent are equal to the lengths of the two segments of the other tangent. That is:
AB = CD
BC = DE
where AB and BC are the lengths of the two segments of one tangent, and CD and DE are the lengths of the two segments of the other tangent.
Using these formulas, we can solve for segment length in various situations involving circles, chords, secants, and tangents.
To convert the degree measure to radian measure, we use the fact that 360 degrees is equal to 2π radians.
Therefore, we can use the following conversion formula:
radian measure = (degree measure × π) / 180
For example:
Convert 45 degrees to radians:
radian measure = (45 degrees × π) / 180
radian measure = (45/180)π
radian measure = π/4
So 45 degrees is equal to π/4 radians.
Convert 120 degrees to radians:
radian measure = (120 degrees × π) / 180
radian measure = (2/3)π
So 120 degrees is equal to (2/3)π radians.
Convert 270 degrees to radians:
radian measure = (270 degrees × π) / 180
radian measure = (3/2)π
So 270 degrees is equal to (3/2)π radians.
Note that radians are a more natural unit for measuring angles in many mathematical contexts, as they relate directly to the arc length of a circle.
To convert the radian measure to degree measure, we use the fact that 180 degrees equal π radians.
Therefore, we can use the following conversion formula:
degree measure = (radian measure × 180) / π
For example:
Convert π/3 radians to degrees:
degree measure = (π/3 radians × 180) / π
degree measure = 60 degrees
So π/3 radians is equal to 60 degrees.
Convert 2π/5 radians to degrees:
degree measure = (2π/5 radians × 180) / π
degree measure = (360/5) degrees
degree measure = 72 degrees
So 2π/5 radians is equal to 72 degrees.
Convert 3π/4 radians to degrees:
degree measure = (3π/4 radians × 180) / π
degree measure = (540/4) degrees
degree measure = 135 degrees
So 3π/4 radians is equal to 135 degrees.
Note that degree measure is commonly used in everyday life and in many technical fields, whereas radian measure is often used in advanced mathematics, physics, and engineering.
Thus,
The segment length and the conversion of radian and degree are given above.
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Use the given information to find the minimum sample size required to estimate an unknown population mean .
22) How many women must be randomly selected to estimate the mean weight of women in one age
group. We want 90% confidence that the sample mean is within 3.4 lb of the population mean, and
the population standard deviation is known to be 25 lb.
A) 145
B) 147
C) 208
D) 148
The minimum sample size required to estimate an unknown population mean is 148. So, the correct option is D) 148.
To find the minimum sample size required to estimate an unknown population mean with 90% confidence, within 3.4 lb of the population mean, and a population standard deviation of 25 lb, follow these steps:
1. Identify the given values:
- Confidence level = 90%
- Margin of error (E) = 3.4 lb
- Population standard deviation (σ) = 25 lb
2. Find the corresponding z-score for the 90% confidence level. Using a standard normal distribution table or calculator, the z-score is 1.645.
3. Use the formula to find the sample size (n):
n = (z * σ / E)^2
n = (1.645 * 25 / 3.4)^2
4. Calculate the sample size:
n ≈ 147.267
Since we cannot have a fraction of a person, round up to the nearest whole number to ensure the required confidence level.
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e) Find the probability that less than 61% of sampled teenagers own smartphones.
(a) Find the mean :The mean μ p is 0.55
(b) Find the standard deviation
The standard deviation σp is 0.0397
please help find problem (e).. i dont know how to do it
The probability that less than 61% of sampled teenagers own smartphones is 0.934 or 93.4%.
To find the probability that less than 61% of sampled teenagers own smartphones, we need to use the standard normal distribution. We first need to standardize the value of 61% using the formula:
z = (x - μ) / σ
where x is the value we want to standardize (in this case, 61%), μ is the mean (0.55), and σ is the standard deviation (0.0397).
Plugging in the values, we get:
z = (0.61 - 0.55) / 0.0397 = 1.511
We can then look up the probability of getting a z-score less than 1.511 in a standard normal distribution table or calculator. The probability is approximately 0.934.
Therefore, the probability that less than 61% of sampled teenagers own smartphones is 0.934 or 93.4%.
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Complete question:
e) Find the probability that less than 61% of sampled teenagers own smartphones.
(a) Find the mean :The mean μ p is 0.55
(b) Find the standard deviation
The standard deviation σp is 0.0397
18. Simplify -4-√-18
Answer:Step 1:
Enter the expression you want to simplify into the editor.
The simplification calculator allows you to take a simple or complex expression and simplify and reduce the expression to it's simplest form. The calculator works for both numbers and expressions containing variables.
Step-by-step explanation: I really hope this helps
Use upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width). (Round your answers to three decimal places. ) y = V 3x upper sum lower sum у 1
To approximate the area of the region defined by[tex]y = √(3x)[/tex] using upper and lower sums, we first divide the interval [0,1] into n subintervals of equal width [tex]Δx = 1/n[/tex]. We then compute the upper and lower sums using the formulae above, and take their average to obtain the approximate area.
To approximate the area of the region defined by the function [tex]y = √(3x)[/tex]using upper and lower sums, we first need to divide the interval of integration [0,1] into subintervals of equal width. Let n be the number of subintervals, then the width of each subinterval is[tex]Δx = 1/n[/tex].
The upper sum is the sum of the areas of rectangles whose heights are taken from the upper endpoints of each subinterval. Specifically, for each i from 1 to n, we compute the height of the rectangle as f(xi), where xi is the upper endpoint of the i-th subinterval.
Upper sum =[tex]Δx [f(x1) + f(x2) + ... + f(xn)], where x1 = 0, x2 = Δx, x3 = 2Δx, ..., xn = (n-1)Δx.[/tex]Similarly, the lower sum is the sum of the areas of rectangles whose heights are taken from the lower endpoints of each subinterval.
Lower sum = [tex]Δx [f(x0) + f(x1) + ... + f(xn-1)][/tex], where[tex]x0 = 0, x1 = Δx, x2 = 2Δx, ..., xn-1 = (n-1)Δx.[/tex] To find the approximate area of the region using upper and lower sums, we simply compute the upper and lower sums using the given number of subintervals, and take their average: Approximate area = (Upper sum + Lower sum)/2.
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Find the probability of exactly one
successes in five trials of a binomial
experiment in which the probability of
success is 5%.
P = [? ]%
Round to the nearest tenth of a percent.
Enter
The probability of exactly one success in the binomial experiment would be 20. 4 %.
How to find the probability ?The probability that there is one success in a binomial probability which has a chance of success of 5 % can be found by the formula :
P ( X = 1) = (5 choose 1) x ( 0.05 ) x (0.95 ) ⁴
= ( 0.05 ) x ( 0. 95 ) ⁴
= 0.05 x 0.8145
= 0.040725
Multiplying both gives:
P(X = 1) = 5 x 0.040725
= 0.203625
In conclusion, the probability of one success is 0.203625 or 20. 4 %.
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A department store has an odd, but logical way of pricing their toys
A doll was $17
A kite was $14
A pair of skates was $24
using this logic, how much would Legos cost?
hint: it has to do with vowels and consonants
I can't figure it out
The cost of Legos, given that this is based on vowels and consonants would be $ 19.
How to find the cost ?The vowels and consonants can be arranged such that:
Doll - 1 vowel (o), 3 consonants ( d , l , l ) - $ 17
Kite - 2 vowels ( i, e ) , 2 consonants ( k, t) - $ 14
Skates - 2 vowels ( a, e ), 4 consonants ( s, k , t , s) - $24
Using this, we can solve for vowels and consonants such that cost per vowel is $2, and the cost per consonant is $5.
The cost of Legos is based on 2 vowels (e, o) and 3 consonants (L, g, s)
= ( 2 x 2 ) + ( 5 x 3 )
= $ 19
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Jonty has a storage container in the shape of a cuboid
Jonty is correct as the original cost of paint is £196 which is less than £200 as said by him.
The area of part of cuboid to be painted will be the sum of all the unpainted areas. So, the area remaining to be painted will be = (2 × 3 × 2.5) + (2 × 3 × 12) + (12 × 2.5)
Remaining area = 15 + 72 + 30
Remaining area = 117 m²
Let us assume the original cost of paint be x. So,
x + 10%x = 26.95
110x = 26.95 × 100
110x = 2695
x = £24.5
Now, number of required tins = total unpainted area/area covered by one tin
Number of required tins = 117/15
Number of required tins = 7.8 tins
Taking it as 8 tins.
Previous cost of tins = 8 × 24.5
Previous cost = £196
Since the original cost is less than £200, Jonty is stating truth.
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The complete question is attached in figure.
Solve this please thank u :)
make it simple to
The missing figures in the diagrams are: 88km 616km², 37.7km 113.14km respectively
What is a circle?A circle is a shape consisting of all points in a plane that are at a given distance from a given point, the center. Equivalently, it is the curve traced out by a point that moves in a plane so that its distance from a given point is constant.
Radis diameter circumference area
S/N r 2r 2πr πr²
Applying the above formulae in each of the questions we have as follows:
1 3 2*3=6 2*3.14*3=18.84 3.14*3*3=28.26 2 3.5 7 2*3.14*3.5=21.98 3.14*3.5*3.5=38.47
3 7.5 15 2*3.14*7.5=47.1ft 3.14*7.5*7.5=176.25
4 14km 28km 2*3.14*14=87.92km 3.14*14*14=615.44km²
5 5mi 10mi 2*3.14*5=31.4mi 3.14*5*5= 78.5mi²
6 2.5cm 5cm 2*3.14*2.5=15.7cm 3.14*2.5*2.5=19.63cm²
7
14 14*2 28 2*22/7*14 22/7*14*14
88km 616km²
6 12 2*22/7*6 22/7*6*6
37.7km 113.14km
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Please help ASAP! I need to finish this TODAY
The school which is a better choice is sea side.
We are given that;
The plot
Now,
If you are interested in a smaller class size, Seaside School is a better choice for you because it has a smaller mean and median class size than Bay Side School. This means that on average and in general, Seaside School has fewer students per class than Bay Side School. Also, Seaside School has a smaller maximum class size than Bay Side School (both have a minimum of zero), so you are less likely to encounter a very large class at Seaside School.
Therefore, by algebra the answer will be sea side.
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Suppose you are using α = 0. 05 to test the claim that μ = 1620 using a P-value. You are given the sample statistics n-35, X_bar=1590 and σ=82. Find the P-value. State the answer only and no additional work. Make sure to use the tables from the book. Do not round the final answer
The P-value is 0.0107 for the sample statistics n-35 and the coefficient of standard deviation is 82.
α = 0. 05
μ = 1620
size (n)= 35
X_bar=1590
σ=82
From the given sample statistics, the test statistics will be calculated as:
t = (X_bar - μ) / (σ / sqrt(n))
t = (1590 - 1620) / (82 / sqrt(35))
t = (-2.5411)
Using the t-distribution table with 34 degrees of freedom, the critical value will be:
t_critical = -1.6909
Here the calculated test statistic is less than the critical value.
P - value = 2*P(-100< t < -1.9720, when df = 34)
P = tcf (-100,-2.4103,34)
P = 0.0107
Therefore we can conclude that the P-value is 0.0107.
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If n = 25, 4 = 20%, M = 10%, and s = 15%,
Use the drop-down menus to complete this sentence that reports the results. (Note: 'XXX' is put in place of the actual numbers so as to not give away what the correct values are for the previous
questions.)
There [8a. Select] a significant reduction in peoples over estimation of the line length, [8b. Select], p [8c. Select], with [8d. Select]
8a.
A. was
B. was not
8b.
A. t(df) = XXX
B. t = XXX with df = XXX
C. t-test with df = XXX
D. M = 10%
8c.
A. < 0.01 two-tailed
B. > 0.01 two-tailed
C. = 0.01
8d.
A. Cohen's d = XXX, M = 10%, 95% CI [XXX, XXX].
B. M = 10%, n = 25, s = 15%.
C. M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].
D. the t-test showing that people did do better after training.
There was a significant reduction in peoples over estimation of the line length, t = XXX with df = XXX, p < 0.01 two-tailed, with M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].
8a. A. was
8b. B. t = XXX with df = XXX
8c. A. < 0.01 two-tailed
8d. C. M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].
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rearrange the formulas to find r
I=Pr + t
The solution of the formula for the variable r is given as follows:
r = (I - t)/P.
How to solve the formula for the variable r?The formula in this problem is defined as follows:
I = Pr + t.
To solve the formula for the variable r, we first must isolate the term with the variable r, as follows:
Pr = I - t.
Then we isolate the variable r applying the division operation, which is the inverse operation to the multiplication, giving the solution as follows:
r = (I - t)/P.
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Answer all boxes and read the questions
The area of the lateral face of cylinder = 75.4 in²
The area of the two bases of the cylinder = 25.13 in²
The total surface area of the cylinder = 100.53 in²
We know that the formula for the surface area of cylinder is:
A = 2πrh + 2πr²
where r is the radius of the cylinder
and h is the height of the cylinder
Here, r = 2 in and h = 6 in
The area of the lateral face of cylinder is given by,
A₁ = 2 × π × r × h
A₁ = 2 × π × 2 × 6
A₁ = 24 × π
A₁ = 75.4 sq. in.
And the area of two base is,
A₂ = 2πr²
A₂ = 2 × π × 2²
A₂ = 8 × π
A₂ = 25.13 sq. in.
The total surface area of cylinder would be,
A = A₁ + A₂
A = 75.4 + 25.13
A = 100.53 sq. in.
Therefore, the required area = 100.53 in²
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Cruz purchased a large pizza for $12.75. It serves 5 people. What is the cost per serving?
$2.55 per serving
$2.60 per serving
$3.15 per serving
$7.55 per serving
If cruz purchased a large pizza for $12.75. It serves 5 people, the cost per serving of the pizza is $2.55. So, correct option is A.
To find the cost per serving of the pizza, we need to divide the total cost of the pizza by the number of servings. In this case, the pizza costs $12.75 and serves 5 people.
Therefore, the cost per serving can be calculated as:
Cost per serving = Total cost of pizza / Number of servings
Cost per serving = $12.75 / 5
Cost per serving = $2.55
So, the cost per serving of the pizza is $2.55.
When working with fractions or dividing quantities, we need to pay attention to the units involved. In this case, the units of the cost and the servings must match for the division to be meaningful.
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Which lists contain only rational numbers? Select all that apply.
The lists that contain only rational numbers is 4/3, -12/13 ,9/4 , -5/7 , 3/4
How can the rational numbersbe known?A rational number can be described as the number which can be expressed in the form of p/q where p and q are integers when writing this number, q must not equal to 0 .
Examples of rational numbers are , however in mathematics, a rational number i can be seen as one that can be expressed as the quotient or fraction which can involves two integers, wherby one will be the numerator p and a non-zero as well as the denominator q.
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please describes in two sentences for each graph if the discrimant is positive, negative, or 0.
1. The discriminant is positive, it has two real solutions
2. The discriminant is zero, it has a real solution
3. The discriminant is negative, it has no real solution
What is the discriminant of a graph?The discriminant of a graph is expressed as the part of the quadratic formula that is found under the square root symbol: b²-4ac.
It describes and gives information on whether there are two solutions, one solution, or no solutions.
It is important to note the following about discriminants;
If the discriminant is zero, then, the equation has real root valuesIf the discriminant is negative, then, the equation has no real root valuesIf the discriminant is positive, then, the equation has two different real root valuesLearn more about discriminants at: https://brainly.com/question/24730520
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Four times a number increased by 25 is 13 less than six times the number. Find the number
Answer:
19
Step-by-step explanation:
Let's call the number we're trying to find "x".
According to the problem:
4x + 25 = 6x - 13
To solve for x, we can start by isolating the x term on one side of the equation. Let's subtract 4x from both sides:
4x + 25 - 4x = 6x - 13 - 4x
25 = 2x - 13
Next, let's add 13 to both sides:
25 + 13 = 2x - 13 + 13
38 = 2x
Finally, we can divide both sides by 2 to solve for x:
38/2 = 2x/2
19 = x
20) As noted on page 332, when the two population means are equal, the estimated standard error for the independent-measures t test provides a measure of how much difference to expect between two sample means. For each of the following situations, assume that u1 = u2 and calculate how much difference should be expected between the two sample means.
One sample has n = 6 scores with SS = 500 and the second sample has n = 12 scores with SS = 524.
One sample has n = 6 scores with SS = 600 and the second sample has n = 12 scores with SS 5 696.
In Part b, the samples have larger variability (bigger SS values) than in Part a, but the sample sizes are unchanged. How does larger variability affect the magnitude of the standard error for the sample mean difference?
We can expect a difference of about 6.67 between the two sample means.
To calculate how much difference to expect between two sample means when the population means are equal, we need to compute the standard error of the difference between means (SED).
The formula for SED in the independent-measures t-test is:
SED = sqrt((s1^2/n1) + (s2^2/n2))
where s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
a) For the first situation, we have:
s1^2 = SS1/(n1-1) = 500/(6-1) = 100
s2^2 = SS2/(n2-1) = 524/(12-1) = 49.45
Plugging these values into the formula, we get:
SED = sqrt((100/6) + (49.45/12)) = 5.76
Therefore, we can expect a difference of about 5.76 between the two sample means.
b) For the second situation, we have:
s1^2 = SS1/(n1-1) = 600/(6-1) = 120
s2^2 = SS2/(n2-1) = 696/(12-1) = 69.6
Plugging these values into the formula, we get:
SED = sqrt((120/6) + (69.6/12)) = 6.67
Therefore, we can expect a difference of about 6.67 between the two sample means.
When the samples have larger variability (bigger SS values), the standard error for the sample mean difference will increase. This is because larger variability means that the scores are more spread out around their respective means, which increases the amount of variability in the difference between the two sample means. In contrast, when the variability is smaller, the scores are more tightly clustered around their means, and the standard error for the sample mean difference will be smaller.
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y=-x^2-4x-4 find y coordinate
in a recent year, a hospital had 4126 births. Find the mean number of births per day, then use that result and the poisson distribution to find the probability that in a day, there are 14 births. Does it appear likely that on any given day, there will be exactly 14 births?
While it is not very likely that there will be exactly 14 births on any given day, it is still possible, and the probability of it happening is about 8.3%.
Let's start by calculating the mean or average number of births per day. To do this, we divide the total number of births in a year (4126) by the number of days in a year. Since there are 365 days in a year, the mean number of births per day is:
4126 / 365 = 11.3
This means that on average, there are about 11 to 12 births per day in this hospital.
In this case, the average rate of occurrence is 11.3 births per day. Using the Poisson distribution formula, we can calculate the probability of having 14 births in a day as follows:
P(X=14) = (e⁻¹¹°³) x (11.3¹⁴) / 14!
where e is the mathematical constant approximately equal to 2.71828, X is the random variable representing the number of births in a day, and ! represents the factorial function.
Using a calculator or a software tool, we get:
P(X=14) = 0.083
This means that the probability of having exactly 14 births in a day is about 8.3%.
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How are the hours spent on homework per day related to grade level in school?
Grade in School Hours spent on Homework (per day)
4 1
6 1.5
8 2.5
10 3
12 3.5
a. The higher the grade level in school, the more hours spent on homework.
b. The more hours spent on homework, the lower the grade in school.
c. The higher the grade level in school, the less hours spent on homework.
d. No relationship.
The requried relation is, the higher the grade level in school, the more hours spent on homework. Option A is correct.
From the given data, we can see that the hours spent on homework per day increase as the grade level in school increases. This suggests that there is a positive correlation between grade level and homework hours.
In general, higher grade levels in school involve more advanced coursework and greater academic expectations, which can require more time and effort outside of class to complete homework and study. Therefore, it is reasonable to expect that students in higher grade levels will spend more hours on homework per day compared to students in lower grade levels.
Thus, the requried relation is, the higher the grade level in school, the more hours spent on homework. Option A is correct.
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DE≅AE ,BA∥CE , CB∥DA and m∠C=65∘
The measure of <BAE is 130 degree.
We have,
m <C = 65
As, opposite sides are parallel then ABCD is a parallelogram.
Then, opposite angles of a parallelogram are congruent
∠ BAD = ∠ BCD = 65°
∠ ADE = ∠ BAD = 65° ( alternate angles )
Since, DE = AE then Δ ADE is isosceles Triangle.
So, ∠ DAE = ∠ ADE = 65°
We can write,
∠ BAE = ∠ BAD + ∠ DAE = 65° + 65° = 130°
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A person paid by the hour works 25 hours a week and makes $539. How much would they make if they work 54 hours? Learn This: Multiply 25 with 539 and 54 Round your answer to 2 decimal places
Therefore, if the person works 54 hours, they would make $1,163.04. Rounded to 2 decimal places, the answer is $1,163.00.
The decimal system employs ten decimal digits, a decimal mark, and a minus sign ("-") for negative quantities when writing numbers. The decimal digits are 0 through 9, with the dot (".") serving as the decimal separator in many (mainly English-speaking) nations and the comma (",") in others.
The fractional portion of the number is represented by the place value that follows the decimal. The number 0.56, for instance, is composed of 5 tenths and 6 hundredths.
We can use proportionality to solve this problem. If the person works 25 hours and makes $539, then their hourly rate is:
$539 ÷ 25 hours = $21.56 per hour
They would make if they work 54 hours, we can multiply their hourly rate by the number of hours worked:
$21.56 per hour × 54 hours = $1,163.04
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