State all possible names for each figure.

Answer:

[tex]z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637[/tex]  

[tex]p_v =P(z>3.637)=0.000138[/tex]  

Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.  

Step-by-step explanation:

Data given and notation

[tex]\bar X_{1}=35621[/tex] represent the mean for sample 1  

[tex]\bar X_{2}=33498[/tex] represent the mean for sample 2  

[tex]\sigma_{1}=4310[/tex] represent the population standard deviation for 1  

[tex]\sigma_{2}=4310[/tex] represent the population standard deviation for 2

[tex]n_{1}=100[/tex] sample size for the group 2  

[tex]n_{2}=120[/tex] sample size for the group 2  

[tex]\alpha=0.01[/tex] Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for US residents (sample 1) is higher than the mean for sample 2, the system of hypothesis would be :  

Null hypothesis:[tex]\mu_{1}-\mu_{2}\leq0[/tex]  

Alternative hypothesis:[tex]\mu_{1} - \mu_{2}> 0[/tex]  

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

[tex]z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

[tex]z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637[/tex]  

P value  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>3.637)=0.000138[/tex]  

Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.  

Answer:

84.9

Step-by-step explanation:

The weighted mean is given by the sum of the products of each grade by its respective weight. If the first four grades correspond to 15% of the final grade each, and the final exam is equivalent to 40% of the final grade, Michael's final grade is:

[tex]G= (73+77+82+86)*0.15+(93*0.4)\\G= 84.9[/tex]

Michael's final weighted mean is 84.9.

Michael's final grade, calculated using the weighted mean formula, taking into consideration the individual weights of each test (15%) and the final exam (40%), is approximately 85.4.

The subject of this problem is weighted mean, which is used quite often in the field of statistics. The formula to calculate the weighted mean is *(w1*x1 + w2*x2 + ... + wn*xn) / (w1 + w2 + ... + wn), where w represents the weights and x represents the values.

In Michael's case, his weights for the test grades are 15% for each test, and 40% for the final exam. Thus, the calculation will be as follows:

(0.15 * 73 + 0.15 * 77 + 0.15 * 82 + 0.15 * 86 + 0.4 * 93) / (0.15 + 0.15 + 0.15 + 0.15 + 0.4)

After calculation, the weighted mean, or the final grade, is approximately 85.4, rounded to one decimal place.

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let P(-2,0) & Q(0,2)

slope = 2-0/0+2

= 1

the slope is 1 which will be 45°

Answer: slope = 1

Step-by-step explanation:

The formula for finding slope is given as :

slope = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]x_{1}[/tex] = -5

[tex]x_{2}[/tex] = 3

[tex]y_{1}[/tex] = -3

[tex]y_{2}[/tex] = 5

substituting the values into the formula , we have :

slope = [tex]\frac{5-(-3)}{3-(-5)}[/tex]

slope = [tex]\frac{5+3}{3+5}[/tex]

slope = [tex]\frac{8}{8}[/tex] = 1

Therefore : the slope of the line is 1

Answer:

Step-by-step explanation:

Let X be the time for any customer to call at any time to make reservation in Nite Time Inn.

Given that X is exponential with mean = 4 minutes

We are to find the probability

(a) 3 minutes or less.

=[tex]P(X\leq 3)=1-e^{-3/4} =0.5276[/tex]

(b) 4 minutes or less.

[tex]=P(X\leq 4)\\=1-e^{-4/4} =0.6321[/tex]

(c) 5 minutes or less.

[tex]=P(X\leq 5)\\=1-e^{-5/4} =0.7135[/tex]

(d) longer than 5 minutes.

=1-P(X≤5) = 0.2865

Answer:

A. x = -4 + 3t, y = 3 - 4t: - 2 lessthanorequalto t lessthanorequalto 3

Step-by-step explanation:

Given that a line in two dimension passes through (-4,3) and (2,-5) oriented in the direction of increasing x

We can write the line equation in parametric form as

[tex]\frac{x-x_1}{x_2-x_1} =\frac{y-y_1}{y_2-y_1} \\\frac{x+4}{6} =\frac{y-3}{-8} =t\\x=-4+6t\\y = 3-8t[/tex]

The values of t when x=-4 is 0 and when x =2 is 1

So t varies from 0 to 1

If instead of t we give t' say which is t +2

then we have

t he parametric equations as

x =-4+3t and y = 3-4t

For  x=-4, t =2 and for x = 2 , t =3

So option A is right.

Answer:

E (Y) = 3

Step-by-step explanation:

If a 4-sided die is being rolled repeatedly; and the odd-numbered rolls (1st 3rd,5th, etc.)

The probability of odd number roll will be, p(T) = [tex]\frac{1}{2}[/tex]

However, on your even-numbered rolls, you are victorious if you get a 3 or 4. Also, the  probability of even number roll, p(U) = [tex]\frac{1}{2}[/tex]

In order to  calculate: E (Y); We can say Y to be the number of times you roll.

We know that;

E (Y) = E ( Y|T ) p(T) + E ( Y|U ) p(U)

Let us calculate E ( Y|T ) and E ( Y|U )

Y|T ≅ geometric = [tex]\frac{1}{4}[/tex]

Y|U ≅ geometric = [tex]\frac{1}{2}[/tex]

also; x ≅ geometric (p)

∴ E (x) = [tex]\frac{1}{p}[/tex]

⇒ [tex]\frac{Y}{T}[/tex] = 4 ; also [tex]\frac{Y}{U}[/tex] = 2

E (Y) = 4 × [tex]\frac{1}{2}[/tex] + 2 ×

        = 2+1

E (Y) = 3

Answer:

e = 0.6342

S = 0.808

γsat = 126.207 lb/ft³

Step-by-step explanation:

Given

VT = 0.320 ft³

WT = 38.9 lb

w = 19.2% = 0.192

Gs = 2.67

then we apply

γ = WT / VT

γ = 38.9 lb / 0.320 ft³

γ = 121.5625 lb/ft³

then we get γdry as follows

γdry = γ / (1 + w)

γdry = 121.5625 lb/ft³ / (1 + 0.192)

γdry = 101.982 lb/ft³

the void ratio (e) can be obtained applying this equation

γdry = Gs*γw / (1 + e)

101.982 = 2.67*62.42 / (1 + e)

⇒ e = 0.6342

We can get the degree of saturation (S) as follows

S*e = Gs*w

S = Gs*w / e

S = 2.67*0.192 / 0.6342

S = 0.808

The saturated unit weight (γsat) will be obtained applying this formula

γsat = (Gs + e)*γw / (1 + e)

γsat = (2.67 + 0.6342)*62.42 lb/ft³/ (1 + 0.6342)

γsat = 126.207 lb/ft³

Answer:

a ) 0.1403604645 and 0.1368

b) 0.3464961 and 0.3485

c) 0.802671982 and 0.8018

Step-by-step explanation:

Y~ B (15,0.45)

Y~ N (15*0.45, 15*0.45*0.55) = Y~ N (6.75, 3.7125)

a) P(Y=5) = 15C5 (0.45)^5 * (0.55)^10 = 0.1403604645

   For normal approximation

P(Y = 5 ) = P ( 4.5 < Y < 5.5 )  ......... continuity correction

Hence,

[tex]P ( 4.5 < Y < 5.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{5.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.16775 < Z < -0.64875 )[/tex]

The probability P ( 4.5 < Y < 5.5 ) = 0.1368

b) P(Y>7) = 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6 + 15C10 (0.45)^10 * (0.55)^5 + 15C11 (0.45)^11 * (0.55)^4 + 15C12 (0.45)^12 * (0.55)^3 + 15C13 (0.45)^13 * (0.55)^2 + 15C14 (0.45)^14 * (0.55) + (0.45)^15

= 0.3464961

  For normal approximation

P(Y > 7 ) = P (Y > 7.5 )  ......... continuity correction

Hence,

[tex]P (Y > 7.5) = P (Z > \frac{7.5-6.75}{\sqrt{3.7125} } ) = P (Z > 0.389249)\\[/tex]

The probability P ( Y>7.5 ) = 0.3485

c) P (4 < Y < 10) = 15C5 (0.45)^5 (0.55)^10 + 15C6 (0.45)^ 6 (0.55)^9 + 15C7 (0.45)^7 (0.55)^8 + 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6

= 0.802671982

For normal approximation

P( 4 < Y < 10 ) = P (4.5< Y < 9.5 )  ......... continuity correction

Hence,

[tex]P ( 4.5 < Y < 9.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{9.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.167748416 < Z < 1.427248064 )[/tex]

The probability P (4.5< Y < 9.5 ) = 0.8018

Answer:

The perimeter is 48 units

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The perimeter of the octagon is the sum of its length sides

so

[tex]P=AB+BC+CD+DE+EF+FG+GH+HA[/tex]

we have

[tex]BC=10\ units\\CD=6\ units\\EF=4\ units\\GH=8\ units[/tex]

substitute

[tex]P=AB+10+6+DE+4+FG+8+HA[/tex]

Combine like terms

[tex]P=AB+DE+FG+HA+28[/tex]

we know that

[tex]BC=DE+FG+HA[/tex] ---> by segment addition postulate

[tex]BC=10\ units[/tex]

so

[tex]DE+FG+HA=10\ units[/tex]

substitute in the expression of perimeter

[tex]P=AB+(DE+FG+HA)+28[/tex]

[tex]P=AB+10+28\\P=AB+38[/tex]

Since

[tex]DC= 6\ units[/tex]

and

[tex]EF = 4\ units[/tex]

The distance between F and line BC must be

[tex]6-4=2\ units[/tex]

so

[tex]AB = HG + 2 = 10\ units[/tex]

substitute

[tex]P=AB+38\\P=10+38=48\ units[/tex]

Answer:

51

Step-by-step explanation:

Substitute the value of a

3(2)³+10(2)²-3(2)-7

3(8)+10(4)-3(2)-7

24+40-6-7

64-13

51

Answer:

1. Linear relation

2. Linear relation

3. Non Linear relation

4. Non Linear relation

Step-by-step explanation:

i) the number of pages a copy machine can print over time, expressed by n = 30m is a linear relation where the value of 'n' is directly proportional to 'm'.

ii) The amount of money an hourly worker makes over time, expressed by a = 15h, is a linear relation where the value of 'a' is directly proportional to 'h'.

iii) The temperature outside at noon each day for a year is a non-linear relation because the temperature will be different every day and not by the same amount.

iv) The height increase of a child over 3 years is also a non-linear relation as the height increase will vary from year to year and not by the same amount.

Answer:

Option d)  Is an alternative method to reducing fractions to the lowest terms.

Step-by-step explanation:

Cancellation Method:

Thus, cancellation method

Option d)  Is an alternative method to reducing fractions to the lowest terms.

The cancellation method is a technique used to reduce fractions to their lowest terms by dividing both the numerator and denominator by their greatest common factor, making option (d) the correct answer.

The cancellation method in mathematics is a technique used to simplify fractions by dividing both the numerator and denominator by their greatest common factor. The goal of this method is to reduce fractions to their lowest terms. In practice, this method involves identifying factors common to both the numerator and the denominator and 'canceling' them out by dividing each by that factor. The cancellation method does not raise fractions to the highest terms, nor does it involve multiplying a number evenly by the numerator and denominator; rather, it simplifies fractions.

The answer to the student's question is option (d), as the cancellation method is indeed an alternative method to reducing fractions to the lowest terms. When working with unit conversions, for example, it's essential to cancel units correctly to ensure the correct units in the final answer. This process requires attention to detail to avoid inverting ratios and ending up with incorrect units.

Answer:there are 16 goats and 22 chickens.

Step-by-step explanation:

Let x represent the number of goats in the field.

Let y represent the number if chicken in the field.

After a quick count, you determine there are 38 heads and 108 feet in the field. A goat has one head. A chicken also has one head. It means that

x + y = 38

A goat has four legs. A chicken has 2 legs. It means that

4x + 2y = 108 - - - - - - - - - - - 1

Substituting x = 38 - y into equation 1, it becomes

4(38 - y) + 2y = 108

152 - 4y + 2y = 108

- 4y + 2y = 108 - 152

- 2y = - 44

y = - 44/ - 2

y = 22

x = 38 - y = 38 - 22

x = 16

Answer:

Step-by-step explanation:

Triangle BCD is a right angle triangle.

From the given right angle triangle

BC represents the hypotenuse of the right angle triangle.

Taking 45 degrees as the reference angle,

BC represents the adjacent side of the right angle triangle.

BD represents the opposite side of the right angle triangle.

To determine BC, we would apply trigonometric ratio

Cos θ = adjacent side/hypotenuse side. Therefore,

Cos 45 = BC/2√2

√2/2 = BC/2√2

BC = 2√2 × √2/2

BC = 2

Answer:

There is a 62% probability that the student will be awarded at least one of the two scholarships.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that the student gets a scolarship from Agency A.

B is the probability that the student gets a scolarship from Agency B.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that the student will get an scolarship from agency A but not from agency B and [tex]A \cap B[/tex] is the probability that the student will get an scolarship from both agencies.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

What is the probability that the student will be awarded at least one of the two scholarships?

This is

[tex]P = a + b + (A \cap B)[/tex]

We have that:

[tex]A = 0.55, B = 0.40[/tex]

If the student is awarded a scholarship from Agency A, the probability that the student will be awarded a scholarship from Agency B is 0.60.

This means that:

[tex]\frac{A \cap B}{A} = 0.6[/tex]

[tex]A \cap B = 0.6A = 0.6*0.55 = 0.33[/tex]

----------

[tex]A = a + (A \cap B)[/tex]

[tex]0.55 = a + 0.33[/tex]

[tex]a = 0.22[/tex]

--------

[tex]B = b + (A \cap B)[/tex]

[tex]0.40 = b + 0.33[/tex]

[tex]b = 0.07[/tex]

Answer:

[tex]P = a + b + (A \cap B) = 0.22 + 0.07 + 0.33 = 0.62[/tex]

There is a 62% probability that the student will be awarded at least one of the two scholarships.

The probability that the student will be awarded at least one of the two scholarships is 0.73.

To calculate the probability that the student will be awarded at least one of the two scholarships, we can use the following formula:

P(at least one scholarship) = 1 - P(no scholarships)

The probability that the student will not be awarded a scholarship from either agency is the product of the probability that the student will not be awarded a scholarship from Agency A and the probability that the student will not be awarded a scholarship from Agency B.

The probability that the student will not be awarded a scholarship from Agency A is

1 - 0.55 = 0.45.

The probability that the student will not be awarded a scholarship from Agency B is

1 - 0.40 = 0.60.

Therefore, the probability that the student will not be awarded a scholarship from either agency is

0.45 * 0.60 = 0.27.

Therefore, the probability that the student will be awarded at least one of the two scholarships is

1 - 0.27 = 0.73.

Another way to calculate this probability is to use the following formula:

P(at least one scholarship) = P(scholarship from Agency A) + P(scholarship from Agency B) - P(scholarship from both agencies)

We already know the probability that the student will be awarded a scholarship from each agency. The probability that the student will be awarded a scholarship from both agencies is the product of the probability that the student will be awarded a scholarship from Agency A and the probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A.

The probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A is 0.60.

Therefore, the probability that the student will be awarded a scholarship from both agencies is

0.55 * 0.60 = 0.33.

Therefore, the probability that the student will be awarded at least one of the two scholarships is

0.55 + 0.40 - 0.33 = 0.73.

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a) The surface area, S, as a function of radius r is given by S = 2πrh + 2πr².

b) The value of S (surface area) approaches infinity as r approaches infinity.

a) The surface area of a closed cylindrical can consists of two parts: the lateral surface area and the two circular base areas.

Let's denote the height of the cylinder as 'h'.

The volume V of the cylinder is given by:

V = πr²h.

We want to express the surface area S in terms of the radius r.

The lateral surface area is given by:

Lateral Surface Area = 2πrh.

Each circular base has an area of πr².

So, the total surface area S is:

S = 2πrh + 2πr².

Hence,  the total surface area S is S = 2πrh + 2πr².

b) As r approaches infinity, let's analyze the behavior of the surface area S:

S = 2πrh + 2πr².

As r gets larger and larger, the term 2πr² dominates the expression.

This is because the term 2πrh (the lateral surface area) is proportional to both 'r' and 'h', whereas the term 2πr² (the circular base areas) is solely proportional to 'r²'.

So, as 'r' approaches infinity, the value of 2πr² will also approach infinity, and this will greatly outweigh the influence of the term 2πrh.

Therefore, the value of S will tend toward infinity as 'r' approaches infinity.

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a) S(r) = 2πr^2 + (2V/r)

b) As r→∞, S(r)→∞, the surface area increases without bound, so it approaches infinity.

a) To find the surface area, S, of a closed cylindrical can as a function of its radius r, you can use the formula for the surface area of a closed cylinder:

S = 2πr^2 + 2πrh

Where:

S is the surface area,

π (pi) is a mathematical constant approximately equal to 3.14159,

r is the radius of the cylinder,

and h is the height (or length) of the cylinder.

However, you mentioned that the volume V is fixed, so we can express the height (h) of the cylinder in terms of its radius (r) and fixed volume (V). The volume of a cylinder is given by:

V = πr^2h

Solving for h:

h = V / (πr^2)

Now, substitute this expression for h into the formula for the surface area:

S = 2πr^2 + 2πr(V / (πr^2))

Simplify:

S = 2πr^2 + 2V/r

So, the surface area, S, as a function of the radius r for a closed cylindrical can with a fixed volume V is:

S(r) = 2πr^2 + 2V/r

b) As r approaches infinity, let's analyze what happens to the value of S(r):

S(r) = 2πr^2 + 2V/r

The first term, 2πr^2, is a quadratic term in r. As r becomes very large (approaching infinity), this term dominates the expression, and S(r) grows without bound. In other words, it approaches infinity.

The second term, 2V/r, is inversely proportional to r. As r becomes very large, this term approaches zero. However, the first term dominates the behavior of S(r), so the overall behavior is dominated by the growth of the quadratic term.

So, as r approaches infinity, the value of S(r) approaches infinity (i.e., S(r) tends toward infinity).

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Answer:

Step-by-step explanation:

The given simultaneous equations are expressed as

y = 2x + 7 - - - - - - - - - -1

y = -2x - 5 - - - - - - - - - - 2

We would apply the method of substitution.

The first step is to equate equation 1 to equation 2. It becomes

2x + 7 = - 2x - 5

Next step is to add 2x to the left hand side and the right hand side of the equation. It becomes

2x + 2x + 7 = - 2x - 2x - 5

4x + 7 = - 5

Next step is to subtract 7 from the left hand side and the right hand side of the equation. It becomes

4x + 7 - 7 = - 5 - 7

4x = - 12

Next step is to divide the left hand side and the right hand side of the equation by 4. It becomes

4x/4 = -12/4

x = - 3

Substituting x = - 3 into equation 1, it becomes

y = 2 × - 3 + 7 = - 6 + 7

y = 1

Answer:

c). Two tailed test

Step-by-step explanation:

The given hypothesis are

Null hypothesis: H0:μ= 1.7

Alternative hypothesis: H1:μ≠ 1.7

The alternative hypothesis demonstrates that  mean number of children are not 1.7 in 2000. This means that mean number of children can be greater than 1.7 or mean number of children can be less than 1.7. Thus, the given alternative hypothesis indicates the two tailed test.

To calculate the resultant force and its direction given two forces at angles, you decompose each force into x and y components, sum these components separately to find the resultant vector, and then use Pythagorean theorem and inverse tangent to find magnitude and direction.

To find the magnitude of the resultant force and the angle it makes with the positive x-axis, given forces at 45 degrees and 30 degrees below the x-axis with magnitudes 28 lb and 12 lb respectively, we break each force into its x and y components. For force a at 45 degrees, the components are 28cos(45) in the x-direction and 28sin(45) in the y-direction. For force b at -30 degrees, the components are 12cos(-30) in the x-direction and 12sin(-30) in the y-direction, since it is below the x-axis.

To find the resultant force (Fres), we add the x-components and y-components separately: Fres,x = 28cos(45) + 12cos(-30) and Fres,y = 28sin(45) + 12sin(-30). The total magnitude is calculated using the Pythagorean theorem: |Fres| = sqrt(Fres,x² + Fres,y²). The angle θ with the positive x-axis is found using the inverse tangent of the y-component over the x-component (θ = atan(Fres,y/Fres,x)).

Final answer:

To find the probability of selecting one ace and one of either a ten, jack, queen, or king from a deck with replacement, calculate the probability for each draw and account for both possible orders of drawing these cards the final probability is 2 * ((4/52) * (16/52)) to account for both orders.

Explanation:

To calculate the probability that one card is an ace and the other is either a ten, jack, queen, or king when two cards are selected randomly from an ordinary deck, we approach this problem considering the combinations and probabilities of each event happening.

First, let's establish the total possibilities. Since we are replacing the card back into the deck after the first draw, the number of possibilities for each draw remains 52. So, the total number of ways to draw any two cards (with replacement) is 52 * 52.

The probability of drawing an ace (P(Ace)) on the first draw is 4/52 because there are 4 aces in a deck of 52 cards.

The probability of drawing a ten, jack, queen, or king (P(10/J/Q/K)) on the second draw is 16/52, as there are 4 of each of these cards in the deck, totaling 16 cards.

To find the probability of both events happening, you multiply the probabilities of each event: P(Ace) * P(10/J/Q/K). This gives you (4/52) * (16/52).

However, the order in which the ace and the 10/J/Q/K are drawn matters, as the ace could also be drawn second. This means we need to calculate the probability again with the ace being drawn second and the 10/J/Q/K card drawn first, which is the same calculation. Therefore, the final probability is 2 * ((4/52) * (16/52)) to account for both orders.

To calculate the probability of drawing an ace followed by a ten, jack, queen, or king, we find the probabilities of each event separately and multiply them since the events are independent, assuming the card is replaced after the draw.

The probability of selecting one ace and another card which is a ten, jack, queen, or king involves understanding the composition of a standard deck and calculating the odds of each draw.

A standard deck has 52 cards with 4 aces and 16 face cards (including the tens). The chance of drawing an ace (event A) is therefore 4/52. For the second draw (event B), with the card being replaced after the first draw, the odds of drawing either a ten, jack, queen, or king from any suit remain identical, since we're back to having a full deck.

To find the overall probability, we calculate the probability of both events occurring, which involves multiplying the probabilities of each individual event, assuming they are independent. This is represented by: P(A and B) = P(A) * P(B).

For the specific example of drawing a four (event A) then a five (event B) when replacing the card back into the deck after each draw, the probability of event A is 1/13 since there are four fours in the deck and the probability of event B is also 1/13 for the same reason. Therefore, the combined probability is (1/13)*(1/13), which simplifies to 1/169.

Answer:

Step-by-step explanation:

-2x + 7x + 5 = 0

-2x +7x = 5

5x = 5

divide both side by 5

5x/5 = 5/5

x = 1

Answer:Mean = $248

Median = $252

Quartile 3 = $265

Mode = $265

range = $69

Standard deviation= 20.126

Coefficient of variation = 8.115

Negatively

Coefficient of skewness = -0.596

Quartile 2 = $252

Step-by-step explanation:

The detailed explanation can be found in the attached pictures

Answer:

C. Total flow

Step-by-step explanation:

The total amount of fluid that has passed a designated point is designated as the total flow.

The flow rate is the rate at which fluid passes through a certain point, not the total amount of fluid.

Laminar and turbulent flow are different types of fluid flow patterns and don't characterize a volume of fluid.

Therefore, the correct answer is C. Total flow.

The formula used is: [tex]d = \frac{C}{ \pi }[/tex]

The diameter is 17.83 feet

Solution:

Given that,

Circumference of pool = C = 56 feet

To find: diameter (d)

The circumference of circle is given as:

[tex]C = 2 \pi r[/tex]

Where "r" is the radius of circle

We know, that diameter is twice the radius

[tex]d = 2r[/tex]

Thus the formula becomes,

[tex]C = \pi d[/tex]

Rearrange for "d"

[tex]d = \frac{C}{ \pi }[/tex]

Substituting the values we get,

[tex]d = \frac{56}{3.14} = 17.83[/tex]

Thus the diameter is 17.83 feet

Answer:

a) 68% of the investments had a return of between 10% and 30%.

b) 32% of investments had a return that was either less than 10% or morethan 30%.

Step-by-step explanation:

We can use the Empirical Rule to solve this question:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean. This also means that 32% of the measures are more than 1 standard deviation from the mean.

95% of the measures are within 2 standard deviation of the mean. This also means that 5% of the measures are more than 2 standard deviations from the mean.

99.7% of the measures are within 3 standard deviations of the mean. This also means that 0.3% of the measures are more than 3 standard deviations from the mean.

In this problem, we have that:

Mean = 20%.

Standard deviation = 10%.

a. What proportion of the investments had a return of between 10% and 30%?

10 is the mean subtracted by 1 one standard deviation

30 is one standard deviation added to the mean.

So 10 and 30 are within 1 standard deviation of the mean. So 68% of the investments had a return of between 10% and 30%.

b. What proportion of investments had a return that was either less than 10% or morethan 30%?

This is the proportion of investments that were farther than one standard deviation of the mean.

By the Empirical Rule, 32% of investments had a return that was either less than 10% or morethan 30%.

Answer:

d = (max distance / max time)t + 0.

Step-by-step explanation:

Earthquake travels 55km of rock in 25 seconds, the relationship between d, the distance traveled in km, and t, the time elapsed in seconds is:

d = (max distance / max time)t + 0

d = (55 km / 25 sec) * t

  = 2.2t

Answer:

2.2km/secs

Step-by-step explanation:

Speed = distance /time

= 55km/25secs

= 2.2km/secs

Answer:

a) [tex] \frac{ds}{dt}= v(t) = 3t^2 -18t +15[/tex]

b) [tex] v(t=3) = 3(3)^2 -18(3) +15=-12[/tex]

c) [tex] t =1s, t=5s[/tex]

d)  [tex] [0,1) \cup (5,\infty)[/tex]

e) [tex] D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46[/tex]

And we take the absolute value on the middle integral because the distance can't be negative.

f) [tex] a(t) = \frac{dv}{dt}= 6t -18[/tex]

g) The particle is speeding up [tex](1,3) \cup (5,\infty)[/tex]

And would be slowing down from [tex][0,1) \cup (3,5)[/tex]

Step-by-step explanation:

For this case we have the following function given:

[tex] f(t) = s = t^3 -9t^2 +15 t[/tex]

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

[tex] \frac{ds}{dt}= v(t) = 3t^2 -18t +15[/tex]

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

[tex] v(t=3) = 3(3)^2 -18(3) +15=-12[/tex]

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

[tex] 3t^2 -18 t +15 =0[/tex]

We can divide both sides of the equation by 3 and we got:

[tex] t^2 -6t +5=0[/tex]

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

[tex] (t-5)*(t-1) =0[/tex]

And for this case we got [tex] t =1s, t=5s[/tex]

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

[tex] t^2 -6t +5 >0[/tex]

[tex] (t-5) *(t-1)>0[/tex]

So then the intervals positive are [tex] [0,1) \cup (5,\infty)[/tex]

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

[tex] D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6[/tex]

And if we replace we got:

[tex] D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46[/tex]

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

[tex] a(t) = \frac{dv}{dt}= 6t -18[/tex]

Part g and h

The particle is speeding up [tex](1,3) \cup (5,\infty)[/tex]

And would be slowing down from [tex][0,1) \cup (3,5)[/tex]

b) The velocity after 3 seconds is -12 feet per second. c) The particle is at rest when t = 1 and t = 5. d) The particle is moving in the positive direction between the critical points of the velocity function. e) The total distance traveled during the first 6 seconds can be found by integrating the absolute value of the velocity function. f) The acceleration at time t is given by the derivative of the velocity function. h) The particle is speeding up when its acceleration is positive and slowing down when its acceleration is negative.

b) To find the velocity after 3 seconds, we need to find the derivative of the function f(t). The derivative of f(t) is v(t), the velocity function. So, v(t) = f'(t), which is equal to 3t^2 - 18t + 15. Now, to find v(3), we substitute t = 3 into the velocity function:

v(3) = 3(3)^2 - 18(3) + 15

= 27 - 54 + 15

= -12 feet per second

c) The particle is at rest when its velocity is zero. So, to find when the particle is at rest, we need to find the time when v(t) = 0:

0 = 3t^2 - 18t + 15

Solving this quadratic equation, we find that the particle is at rest when t = 1 and t = 5

d) The particle is moving in the positive direction when its velocity is positive. So, we need to find the time intervals when v(t) > 0. We can do this by finding the critical points of the velocity function and determining the sign of v(t) in between those critical points. By analyzing the sign of v(t), we can determine the intervals when the particle is moving in the positive direction.

(e) To find the total distance traveled during the first 6 seconds, we need to find the definite integral of the absolute value of the velocity function from 0 to 6:

Distance = ∫06 |v(t)| dt

(f) The derivative of the velocity function v(t) gives us the acceleration function a(t), which is the rate of change of velocity. So, a(t) = v'(t), which is equal to 6t - 18.

(h) The particle is speeding up when its acceleration is positive, and slowing down when its acceleration is negative. So, we need to find the intervals when a(t) > 0 and a(t) < 0 to determine when the particle is speeding up and when it is slowing down.

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Answer:

See the proof below.

Step-by-step explanation:

Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."

Solution to the problem

For this case we can assume that we have N independent variables [tex]X_i[/tex] with the following distribution:

[tex] X_i Bin (1,p) = Be(p)[/tex] bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:

[tex] Z = \sum_{i=1}^N X_i[/tex]

From the info given we know that [tex] N \sim Bin (M,q) [/tex]

We need to proof that [tex] Z \sim Bin (M, pq)[/tex] by the definition of binomial random variable then we need to show that:

[tex] E(Z) = Mpq[/tex]

[tex] Var (Z) = Mpq(1-pq)[/tex]

The deduction is based on the definition of independent random variables, we can do this:

[tex] E(Z) = E(N) E(X) = Mq (p)= Mpq[/tex]

And for the variance of Z we can do this:

[tex] Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2 [/tex]

[tex] Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2[/tex]

And if we take common factor [tex]Mpq [/tex] we got:

[tex] Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq][/tex]

And as we can see then we can conclude that   [tex] Z \sim Bin (M, pq)[/tex]

To show that X has a binomial distribution with parameters p*n and q*m, we can formulate X as a random sum of indicator random variables Zi = 1 if the i-th trial is a success and 0 otherwise.

Let X be a binomial random variable with parameters p and n, and let Y be a binomial random variable with parameters q and m. To show that X has a binomial distribution with parameters p*n and q*m, we can formulate X as a random sum of indicator random variables Z1, Z2, ..., Zn, where Zi = 1 if the i-th trial is a success and 0 otherwise. Then X can be expressed as X = Z1 + Z2 + ... + Zn. Since each Zi is independent and follows a Bernoulli distribution with parameter p, the sum of n independent Bernoulli random variables is a binomial random variable with parameters p and n. Therefore, X has a binomial distribution with parameters p*n and q*m.

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Answer:

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

Step-by-step explanation:

For this case our variable of interest is the average parking time of its students. And we have 280 values for these times. So then the variable of interest is quantitative.

Which of the following graphs should not be used to display information concerning the students parking times?

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

b.Stem-and-leaf display

That incorrect since the Stem and leaf plot is useful when we want to plot quantitative data.

c. Histogram

That incorrect since the Histogram is ideal when we want to plot quantitative data and analyze the distribution of the data.

d. Box plot

That incorrect since the Box plot is ideal and useful when we want to plot quantitative data and see central tendency measures.

Final answer:

A pie chart should not be used to display the average parking times of students because it cannot effectively represent the distribution or variability of parking times, unlike a histogram, box plot, or stem-and-leaf display.

Explanation:

The question revolves around determining which graph would be inappropriate for displaying the average parking times of students at a university. The options are: a Pie chart, a Stem-and-leaf display, a Histogram, and a Box plot. To assess the suitability of each graph, we need to understand what type of data we have and what information we wish to convey with the graph.

A pie chart is typically used to show parts of a whole. However, in this scenario, we're interested in the distribution of parking times, not how a single parking time compares as a fraction of the total. Therefore, a pie chart would not effectively represent the variability or distribution of parking times.

Both a histogram and a box plot are well-suited for displaying the distribution of quantitative data such as parking times, making them good choices for this context. A stem-and-leaf display is also a valid option, especially for providing a quick visual summary that includes actual data points, which can be useful for identifying specific patterns or outliers.

In conclusion, the pie chart is the graph that should not be used to display information concerning the students' parking times. It simply doesn't align with the objective of analyzing the distribution or average of the parking times, which is best visualized through the other mentioned graphical methods.