A cube of side 5 is built with black and white cubes of side 1, so that the cubes next to each other have different colors and the corner cubes are black, as shown in the figure. How many white cubes were used?

Answer: 5

Step-by-step explanation:

Answer:17

Step-by-step explanation:

The median of this data set is 17, since if you cross one # off of both sides, you will eventually get to the middle fo the data set, pointing to 17.

Answer:

Andre's: Min = 18, Q1 = 23.5, Median = 28.5, Q3 = 31.5, Max = 35, IQR = 8

Lin's: Min = 15, Q1 = 19, Median = 21.5, Q3 = 24, Max = 33, IQR = 5

Noah's: Min = 10, Q1 = 12.5, Median = 19, Q3 = 22.5, Max = 25, IQR = 10

Step-by-step explanation:

Determine if the weights across all types of chocolate are statistically significantly different from one another using a significance level of 0.05. To do this, we'll use an ANOVA (Analysis of Variance) test. Here are the steps to perform the test:

1. Organize the data: Group the weights of each type of chocolate (peanuts, almonds, macadamia nuts, and no nuts) separately.

2. Calculate the means: Find the mean weight for each group and the overall mean weight for all 20 pieces of chocolate.

3. Calculate the Sum of Squares Between (SSB) and Sum of Squares Within (SSW): SSB represents the variation between groups, and SSW represents the variation within each group.

4. Calculate the Mean Squares Between (MSB) and Mean Squares Within (MSW): Divide SSB by the degrees of freedom between groups (k-1, where k is the number of groups), and divide SSW by the degrees of freedom within groups (N-k, where N is the total number of samples).

5. Calculate the F statistic: Divide MSB by MSW.

6. Determine the critical F value: Using an F distribution table, find the critical F value corresponding to a significance level of 0.05 and the degrees of freedom between and within groups.

7. Compare the calculated F statistic to the critical F value: If the calculated F statistic is greater than the critical F value, the difference in weights across the types of chocolate is considered statistically significant.

If you follow these steps with the provided weight data, you'll be able to determine if the differences in chocolate weights are statistically significant at a 0.05 significance level.

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To show as much work as possible and simplify the expression 3(5-7), we first need to simplify the expression inside the parentheses. After simplification, we get the answer as -6.

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Answer: $13.5

Step-by-step explanation:

1. 15 x 10 =150

2. 150/100=1.5

3. 15.00 - 1.50= 13.5

Answer:

-----------------------

The perimeter is the sum of three side lengths:

Answer:

a) It is not possible to find the mean because these are words, not numbers.

b) If we put these words in alphabetical order, we have:

blue, blue, green, purple, purple, purple, red, red

The median word here is purple.

c) It is possible to find the mode, which in this case is the word that appears the most times in this list. That word is purple, which appears three times.

The correct solution to the given boundary value problem (D^2 + 4^2)y = 0, with y(0) = 0 and y(phi/8) = 1, is d) y = sin 4x.

This can be found by using the value problem characteristic equation of the differential equation, which is r^2 + 16 = 0. Solving for r, we get r = +/- 4i. Therefore, the general solution is y(x) = c1 sin 4x + c2 cos 4x.

To find the values of c1 and c2, we use the boundary conditions. First, we have y(0) = 0, which gives c2 = 0. Then, we have y(phi/8) = 1, which gives c1 = 1/4. Thus, the final solution is y(x) = (1/4) sin 4x.

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The table, we cannot determine the probability of a student participating in exactly two activities.

The probability distribution table is not provided in the question, but assuming that it is a valid probability distribution, we can use it to find the probability that a student participates in exactly two activities.

Let X be the random variable representing the number of extracurricular activities a college freshman participates in, and let p(x) denote the probability of X taking the value x.

Then, we want to find p(2), the probability that a student participates in exactly two activities. This can be obtained from the probability distribution table.

Without the table, we cannot determine the probability of a student participating in exactly two activities.

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Answer:

[tex] \sqrt{175 {x}^{2} {y}^{3} } = \sqrt{175} \sqrt{ {x}^{2} } \sqrt{ {y}^{3} } = \sqrt{25} \sqrt{ {x}^{2} } \sqrt{ {y}^{2} } \sqrt{7} \sqrt{y} = 5xy \sqrt{7y} [/tex]

We can expect the longest 20% of these scorpions to be above a length of approximately 2.0272 inches.

To find the value above which we could expect the longest 20% of these scorpions, we need to use the z-score formula. First, we need to find the z-score that corresponds to the 80th percentile, which is the complement of the top 20%. Using a standard normal distribution table or calculator, we find that the z-score corresponding to the 80th percentile is 0.84.

Next, we use the formula z = (x - mu) / sigma, where z is the z-score, x is the value we are trying to find, mu is the mean, and sigma is the standard deviation. We plug in the given values and solve for x:

0.84 = (x - 1.96) / 0.08

0.0672 = x - 1.96

x = 2.0272

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The mean of the data-set in this problem is given as follows:

41.6 inches.

The mean of a data-set is given by the sum of all observations in the data-set divided by the number of observations, which is also called the cardinality of the data-set.

Considering the stem-and-leaf plot, the observations are given as follows:

20, 32, 34, 36, 40, 42, 44, 48, 55, 65.

The sum of the observations is given as follows:

20 + 32 + 34 + 36 + 40 + 42 + 44 + 48 + 55 + 65 = 416 inches.

There are 10 observations, hence the mean is given as follows:

416/10 = 41.6 inches.

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The rectangle which should go in position I is rectangle A.

We are given that;

The rectangles A,B,C and D with numbers

Now,

To take the same the number of side

If we take A on 1 place

F will be on second place

And  B will be on 4th place

Therefore, by algebra the answer will be rectangle A.

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There are 4 aces and 4 nights in a standard 52-card deck. So, the total number of cards that can be considered as a successful outcome is 8. Therefore, the probability of being dealt an ace or an 8 is 8/52 or 2/13. To find the probability of being dealt an Ace or an 8, follow these steps:

1. Identify the total number of cards in the deck: There are 52 cards in a standard deck.

2. Determine the number of Aces and 8s in the deck: There are 4 Aces and 4 eights, totaling 8 cards (4 Aces + 4 eights).

3. Calculate the probability: Divide the number of desired outcomes (Aces and 8s) by the total number of cards in the deck.

Probability = (Number of Aces and 8s) / (Total number of cards)
Probability = 8 / 52

4. Simplify the fraction: 8/52 can be simplified to 2/13.

So, the probability of being dealt an Ace or an 8 from a standard 52-card deck is 2/13.

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(i) c = 1/2 and the joint pdf is f(x, y) = (x+y)/2 over the unit square.

(ii) 1/16

(iii) 1/9

iv)  5/3

(v) E(X+Y) = 5/6.

we have,

i.

In order for f(x, y) to be a valid joint pdf, it must satisfy two conditions:

It must be non-negative for all (x,y)

The integral over the entire support must equal 1.

To satisfy the first condition, we need c(x+y) to be non-negative.

This is true as long as c is non-negative and x+y is non-negative over the support, which is the unit square [0,1]x[0,1]. Since x and y are both non-negative over the unit square, we need c to be non-negative as well.

To satisfy the second condition, we integrate f(x, y) over the unit square and set it equal to 1:

1 = ∫∫ f(x, y) dx dy

 = ∫∫ c(x+y) dx dy

 = c ∫∫ (x+y) dx dy

 = c [∫∫ x dx dy + ∫∫ y dx dy]

 = c [∫ 0^1 ∫ 0^1 x dx dy + ∫ 0^1 ∫ 0^1 y dx dy]

 = c [(1/2) + (1/2)]

 = c

ii.

P(X < 0.5, Y < 0.5) can be found by integrating the joint pdf over the region where X < 0.5 and Y < 0.5:

P(X < 0.5, Y < 0.5) = ∫ 0^0.5 ∫ 0^0.5 (x+y)/2 dy dx

                    = ∫ 0^0.5 [(xy/2) + (y^2/4)]_0^0.5 dx

                    = ∫ 0^0.5 [(x/4) + (1/16)] dx

                    = [(x^2/8) + (x/16)]_0^0.5

                    = (1/32) + (1/32)

                    = 1/16

iii.

P(Y<X) can be found by integrating the joint pdf over the region where

Y < X:

P(Y < X) = ∫ 0^1 ∫ 0^x (x+y)/2 dy dx

        = ∫ 0^1 [(xy/2) + (y^2/4)]_0^x dx

        = ∫ 0^1 [(x^3/6) + (x^3/12)] dx

        = (1/9)

iv.

P(X+Y) < 5 can be found by integrating the joint pdf over the region where X+Y < 5:

P(X+Y < 5) = ∫ 0^1 ∫ 0^(5-x) (x+y)/2 dy dx

          = ∫ 0^1 [(xy/2) + (y^2/4)]_0^(5-x) dx

          = ∫ 0^1 [(x(5-x)/2) + ((5-x)^2/8)] dx

          = 5/3

v.

The expected value of XY can be found by integrating the product xy times the joint pdf over the entire support:

E(XY) = ∫∫ xy f(x, y) dx dy

E(XY) = ∫∫ xy (x+y)/2 dx dy

= ∫∫ (x^2y + xy^2)/2 dx dy

= ∫ 0^1 ∫ 0^1 (x^2y + xy^2)/2 dx dy

= ∫ 0^1 [(x^3*y/3) + (xy^3/6)]_0^1 dy

= ∫ 0^1 [(y/3) + (y/6)] dy

= 1/4

The expected value of X+Y can be found by integrating the sum (x+y) times the joint pdf over the entire support:

E(X+Y) = ∫∫ (x+y) f(x, y) dx dy

      = ∫∫ (x+y) (x+y)/2 dx dy

      = ∫∫ [(x^2+2xy+y^2)/2] dx dy

      = ∫ 0^1 ∫ 0^1 [(x^2+2xy+y^2)/2] dx dy

      = ∫ 0^1 [(x^3/3) + (xy^2/2) + (y^3/3)]_0^1 dy

      = ∫ 0^1 [(1/3) + (y/2) + (y^2/3)] dy

      = 5/6

Thus,

(i) c = 1/2 and the joint pdf is f(x, y) = (x+y)/2 over the unit square.

(ii) 1/16

(iii) 1/9

iv)  5/3

(v) E(X+Y) = 5/6.

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The solution to the product of the given equation is:

-4x² - 3x + 1

A linear equation is defined as an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax + B = 0. Here, x is a variable, A is a coefficient and B is constant.

Thus, looking at the given equation, we have:

(x + 1) * (-4x + 1)

Expanding this gives:

-4x² - 4x + x + 1

-4x² - 3x + 1

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Answer:

15p - 24q +8

Step-by-step explanation:

Answer:

The simplified result of the expression is 0.63.

Step-by-step explanation:

The equivalent expression is (5x + 2)(x + 1)

Algebraic expressions are mathematical expressions that comprises of variables, terms, coefficients, factors and constants.

Also, note that equivalent expressions are defined as expressions having the same solution but differ in the arrangement of its variables.

From the information given, we have that;

5x²+2=-7x

Put into standard form

5x² + 7x + 2 = 0

To solve the quadratic equation, we have to find the pair factors of the product of 5 and 2 that sum up to 7, we have;

Substitute the values

5x² + 5x + 2x + 2 = 0

group in pairs

(5x² + 5x) + ( 2x + 2 ) = 0

factorize the expression

5x(x + 1) + 2(x + 1) = 0

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Wait just ignore this.

Step-by-step explanation:

CI = p(r/100 + 1)^t - p

136.85 = 320(r/100 +1)^4 - 320

320(r/100 + 1)^4 = 456.85

(r/100 + 1)^4 = 456.85/320 = 1.42765625

(r/100+1) = 4th root of 1.42765625 = 1. 093089979

R/100 = 0.93089979

R = 93.0 %

Renting 0 trucks the Marginal Revenue (MR) = Not applicable, and Average Revenue (AR) = Not applicable. Renting 1 truck the Marginal Revenue (MR) = $P (since it's the additional revenue gained from renting 1 truck), Average Revenue (AR) = Total Revenue / Quantity = P / 1 = $P.

Renting 2 trucks Marginal Revenue (MR) = ($2P - $P) = $P (since it's the additional revenue gained from renting the second truck), Average Revenue (AR) = Total Revenue / Quantity = 2P / 2 = $P. Renting 3 trucks Marginal Revenue (MR) = ($3P - $2P) = $P (since it's the additional revenue gained from renting the third truck), Average Revenue (AR) = Total Revenue / Quantity = 3P / 3 = $P.

To help you with your question, we need to know the rental price per truck and the costs associated with renting these trucks. Since this information is not provided, I will assume a rental price of P dollars per truck. Based on this assumption, we can calculate total, marginal, and average revenue for Sendit when renting 0, 1, 2, or 3 trucks during the move-in week.

1. Renting 0 trucks:
Total Revenue (TR) = 0 * P = $0
Marginal Revenue (MR) = Not applicable
Average Revenue (AR) = Not applicable

2. Renting 1 truck:
Total Revenue (TR) = 1 * P = $P
Marginal Revenue (MR) = $P (since it's the additional revenue gained from renting 1 truck)
Average Revenue (AR) = Total Revenue / Quantity = P / 1 = $P

3. Renting 2 trucks:
Total Revenue (TR) = 2 * P = $2P
Marginal Revenue (MR) = ($2P - $P) = $P (since it's the additional revenue gained from renting the second truck)
Average Revenue (AR) = Total Revenue / Quantity = 2P / 2 = $P

4. Renting 3 trucks:
Total Revenue (TR) = 3 * P = $3P
Marginal Revenue (MR) = ($3P - $2P) = $P (since it's the additional revenue gained from renting the third truck)
Average Revenue (AR) = Total Revenue / Quantity = 3P / 3 = $P

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Answer: We can factor the given polynomial as follows:

x^8 + 3x^4 - 4 = (x^4 - 1)(x^4 + 4)

= (x^2 - 1)(x^2 + 1)(x^2 - 2x + 1)(x^2 + 2x + 1)

The four factors on the right-hand side are all monic polynomials with integer coefficients that cannot be factored further over the integers. Therefore, we have k = 4, and we can compute p_1(1) + p_2(1) + p_3(1) + p_4(1) as follows:

p_1(1) + p_2(1) + p_3(1) + p_4(1) = (1^2 - 1) + (1^2 + 1) + (1^2 - 2(1) + 1) + (1^2 + 2(1) + 1)

= 0 + 2 + 0 + 6

= 8

Therefore, p_1(1) + p_2(1) + p_3(1) + p_4(1) = 8.

Step-by-step explanation:

The length of the missing sides of the two quadrilaterals are listed below:

In this problem we must determine the length of missing sides in two quadrilaterals, this can be done with the help of Pythagorean theorem and properties for special right triangles:

r = √(x² + y²)

45 - 90 - 45 right triangle

r = √2 · x = √2 · y

Where:

Now we proceed to determine the missing sides for each case:

a = √[(6 - 3)² + 4²]

a = √(3² + 4²)

a = √25

a = 5

Case 2

z = √[(22 - 4√2 - 15)² + 4²]

z = √[(7 - 4√2)² + 4²]

z = 4.219

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Hi! To answer your question, let's analyze the complex function f(z) given by f(z) = 1 + cos(az)/(z^2).

First, we need to identify the poles of the function. A pole occurs when the denominator of the function is zero. In this case, the poles are at z = 0. However, the order of the pole is determined by the number of times the denominator vanishes, which is given by the exponent of z in the denominator. Here, the exponent is 2, so the order of the pole is 2.

Now, let's find the residue of complex function f(z) at the pole z = 0. To do this, we can apply the residue formula for a second-order pole:

Res[f(z), z = 0] = lim (z -> 0) [(z^2 * (1 + cos(az)))/(z^2)]'

where ' denotes the first derivative with respect to z.

First, let's find the derivative:

d(1 + cos(az))/dz = -a * sin(az)

Now, substitute this back into the residue formula:

Res[f(z), z = 0] = lim (z -> 0) [z^2 * (-a * sin(az))]

Since sin(0) = 0, the limit evaluates to 0. Therefore, the residue of f(z) at the pole z = 0 is 0.

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a) The system of equations that represents the scenario is given as follows:

b) The costs are given as follows:

The variables for the system of equations are defined as follows:

Castel spent $115 on 5 rose bushes and 12 geraniums, hence:

5x + 12y = 115.

Kail spent $94 on 10 rose bushes and 8 geraniums, hence:

10x + 8y = 94

Then the system is defined as follows:

Multiplying the first equation by 2 and subtracting by the second, we have that the value of y is obtained as follows:

24y - 8y = 230 - 94

16y = 136

y = 136/16

y = 8.5.

Then the value of x is obtained as follows:

5x + 12(8.5) = 115

x = (115 - 12 x 8.5)/5

x = 2.6.

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Answer:

C

Step-by-step explanation:

y=-2x+4

2x+y=4

2x-2x+4=4

4=4

Y follows a negative binomial distribution with parameters r = 3 and p = 0.97.

The moment generating function (MGF) of a random variable X is defined as Mx(t) = E(e^(tX)).

(a) The given MGF is 0.03 Mx(0) t< - log 0.97 1 -0.97e^(tX)

The MGF of the geometric distribution with parameter p is given by Mx(t) = E(e^(tX)) = Σ [p(1-p)^(k-1)]e^(tk), where the sum is taken over all non-negative integers k.

Comparing this with the given MGF, we can see that p = 0.97. Therefore, X follows a geometric distribution with parameter p = 0.97.

(b) Let Y = X1 + X2 + X3, where X1, X3, and X3 are independent and identically distributed geometric random variables with parameter p = 0.97.

The MGF of Y can be obtained as follows:

My(t) = E(e^(tY)) = E(e^(t(X1 + X2 + X3))) = E(e^(tX1) * e^(tX2) * e^(tX3))

= Mx(t)^3, since X1, X2, and X3 are independent and identically distributed with the same MGF

Substituting the given MGF of X, we get:

My(t) = (0.03 Mx(0) t< - log 0.97 1 -0.97e^(t))^3

Therefore, Y follows a negative binomial distribution with parameters r = 3 and p = 0.97.

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The series ∑n-1 to ∞ [tex](-1)^{n-1} )/\frac{\sqrt(n}{n+5} }[/tex] is convergence (C).

The given series is:

∑n-1 to ∞ [tex](-1)^{n-1} )/\frac{\sqrt(n}{n+5} }[/tex]

To determine if the series converges or diverges, we can use the alternating series test. The alternating series test states that if a series has alternating terms that decrease in absolute value and converge to zero, then the series converges.

In this series, the terms alternate in sign and decrease in absolute value, since the denominator (n) increases as n increases. Also, as n approaches infinity, the term [tex](-1)^{n-1}[/tex]oscillates between 1 and -1, but does not converge to a specific value. However, the absolute value of the term 1/n approaches 0 as n approaches infinity.

Therefore, by the alternating series test, the given series converges. The answer is C (convergence).

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The principal that will grow to $5,100 in two years, eight months at 4.3% compounded semi-annually is approximately $4,568.20.

To find the principal that will grow to $5,100 in two years and eight months at 4.3% compounded semi-annually, you can use the formula for compound interest:

A = P(1 + r/n)^(nt)

where:
A = final amount ($5,100)
P = principal amount (what we're trying to find)
r = annual interest rate (4.3% or 0.043)
n = number of times interest is compounded per year (semi-annually, so 2)
t = time in years (2 years and 8 months or 2.67 years)

First, plug in the values:

$5,100 = P(1 + 0.043/2)^(2*2.67)

Next, solve for P:

P = $5,100 / (1 + 0.043/2)^(2*2.67)

P = $5,100 / (1.0215)^(5.34)

P = $5,100 / 1.11726707

P ≈ $4,568.20

The principal that will grow to $5,100 in two years, eight months at 4.3% compounded semi-annually is approximately $4,568.20.

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