The net of a triangular pyramid is shown.

The net of a triangular pyramid.
[Not drawn to scale]


Each triangle in the net has a base length that measures 6 inches and a height that measures 4 inches. What is the surface area of the pyramid that can be formed from this net?
12 inches squared
24 inches squared
36 inches squared
48 inches square

Answer:

4. False

5. False

Step-by-step explanation:

4. Plug in 9 for x in the inequality:

13 - 9 < 4

4 < 4

We see that 4 is actually not less than 4, but is equal. So, this inequality is FALSE.

5. Plug in 20 for x in the inequality:

45 < 2 * 20 - 5

45 < 40 - 5

45 < 35

Obviously, 45 is larger than 35, so this is also FALSE.

Hope this helps!

Answer:

Both false

Step-by-step explanation:

4.

Substitute 9 in for x in the inequality

13 - x< 4

13 - 9< 4

Subtract

4< 4

4 is not less than 4, but is equal, so this is false.

5.

Substitute 20 in for x in the inequality

45 < 2x - 5

45 < 2(20) - 5

Multiply

45 < 40 - 5

Subtract

45 < 35

Since 35 is not greater than 45, this is false.

Answer:

The store sold 120 ounces fewer ounces of cashew than walnut.

The store sold 11.25 pounds of cashews and 18.75 pounds of walnuts.

To find out how many pounds of each type of nut were sold, we first need to multiply the number of cans sold by the weight of each can. This gives us the total weight in ounces. Then, we convert from ounces to pounds.

For the cashews: 30 cans * 6 ounces/can = 180 ounces. 180 ounces / 16 ounces/pound = 11.25 pounds of cashews.

For the walnuts: 30 cans * 10 ounces/can = 300 ounces. 300 ounces / 16 ounces/pound = 18.75 pounds of walnuts.

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Answer:

1/105 m³

Step-by-step explanation:

the volume of a rectangular prism = base area x height

1/3 x 1/5 x 1/7 = 1/105 m³

Answer:

1/105m^3

Step-by-step explanation:

I did the question and it was correct

Answer:

1.19% probability that fewer than 101 commuters drive to work alone

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]p = 0.75, n = 150[/tex]

[tex]\mu = E(X) = 150*0.75 = 112.5[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150*0.75*0.25} = 5.3[/tex]

(a) What is the probability that fewer than 101 commuters drive to work alone

Using continuity corretion, this is [tex]P(X < 101-0.5) = P(X < 100.5)[/tex], which is the pvalue of Z when X = 100.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{100.5 - 112.5}{5.3}[/tex]

[tex]Z = -2.26[/tex]

[tex]Z = -2.26[/tex] has a pvalue of 0.0119

1.19% probability that fewer than 101 commuters drive to work alone

The probability that fewer than 101 U.S. commuters drive to work alone, based on a 75% solo driving rate, is calculated using the binomial probability formula, resulting in the answer.

To solve this problem, we can use the binomial probability formula, as this is a binomial distribution (success/failure) with a known probability of success.

The formula for the probability mass function of a binomial distribution is:

[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k} \][/tex]

where:

-[tex]\( n \)[/tex] is the number of trials (sample size),

- [tex]\( k \)[/tex]  is the number of successful outcomes,

- [tex]\( p \)[/tex]is the probability of success on a single trial.

In this case, [tex]\( n = 150 \)[/tex] (number of commuters),[tex]\( p = 0.75 \)[/tex] (probability of driving alone), and we want to find the probability that fewer than 101 commuters drive alone (\( k < 101 \)).

[tex]\[ P(X < 101) = P(X \leq 100) = \sum_{k=0}^{100} \binom{150}{k} \cdot 0.75^k \cdot (1 - 0.75)^{150 - k} \][/tex]

Now, we can use a calculator or statistical software to compute this probability. Keep in mind that the binomial coefficient[tex]\(\binom{n}{k}\)[/tex] is the number of ways to choose \(k\) successes from \(n\) trials and can be calculated as[tex]\(\frac{n!}{k! \cdot (n - k)!}\).[/tex]

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Answer:

63.6 square cm

Step-by-step explanation:

[tex]d = 9 \: cm \implies \: r = 4.5 \: cm \\ area \: of \: circle \\ = \pi {r}^{2} \\ = 3.14 \times ( {4.5})^{2} \\ = 3.14 \times 20.25 \\ = 63.585 \\ = 63.6 \: {cm}^{2} \\ [/tex]

Answer:

We conclude that the mean temperature of humans is more than or equal to 98.6°F.

Step-by-step explanation:

We are given that it has long been stated that the mean temperature of humans is 98.6°F. ​However, two researchers currently involved in the subject thought that the mean temperature of humans is less than 98.6°F.

They measured the temperatures of 44 healthy adults 1 to 4 times daily for 3​ days, obtaining 200 measurements. The sample data resulted in a sample mean of 98.3°F and a sample standard deviation of 1°F.

Let [tex]\mu[/tex] = true mean temperature of humans.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 98.6°F   {means that the mean temperature of humans is more than or equal to 98.6°F}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 98.6°F   {means that the mean temperature of humans is less than 98.6°F}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean temperature of 44 adults = 98.3°F

             s = sample standard deviation = 1°F

             n = sample of healthy adults = 44

So, test statistics  =   [tex]\frac{98.3-98.6}{\frac{1}{\sqrt{43} } }[/tex]  ~ [tex]t_4_3[/tex]

                               =  -1.967

Hence, the value of test statistics is -1.967.

Now, P-value of the test statistics is given by;

      P-value = P( [tex]t_4_3[/tex] < -1.967) = 0.029 or 2.9%

Now, here the P-value is 0.029 which is clearly higher than the level of significance of 0.01, so we will not reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean temperature of humans is more than or equal to 98.6°F.

It would actually depend on the way that you want your result. It can be in h, min, s

For example, in hours it would be 5.5h

in minutes, 330min

in seconds, 19800s

She should write 5.5 hours on her time card.

To write 5 hours and 30 minutes on her time card, she can simply write 5.5 hours. This is because 30 minutes is half of an hour, so it can be represented as 0.5 hours.

Answer:

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constructed by cutting four equal-sized surface corners from an 8-inch by 11.5 inch sheet of cardboard and folding up the sides.

1. Determine a function that relates the total surface area, s, (measured in square inches) of the open box to the size of the square cutout x (measured in inches).

2. What is the domain and range of the function s?

3. What is the surface area when a 1" x 1" square is cut out?

4. What size square cutout will result in a surface area of 20 in?

5. What is the surface area of the box when the volume is maximized? (Calculator)

Step-by-step explanation:

Answer:

75%

Step-by-step explanation:

Final answer:

To write the linear equation, let x represent the number of Beats Roses sold and y represent the number of Chocolate Hearts sold. The equation is 5x + 2.50y = 250.

Explanation:

To write a linear equation in standard form for this scenario, we need to define our variables. Let x represent the number of Beats Roses sold and y represent the number of Chocolate Hearts sold.

Based on the given information, we know that the price of Beats Roses is $5, and the price of Chocolate Hearts is $2.50.

The total amount of money Matt made at the end of the dance is $250. Using this information, we can write the equation:

5x + 2.50y = 250

Therefore, the linear equation in standard form for this scenario is 5x + 2.50y = 250.

To solve the equation 2 sin(2θ) = 1, find the angles θ where sin(2θ) equals 1/2. The solutions for θ are π/12 + kπ, where k is any integer.

To solve the equation 2 sin(2θ) = 1, we can divide both sides of the equation by 2 to get sin(2θ) = 1/2. Since the range of sin function is -1 to 1, we need to find the angles θ where sin(2θ) equals 1/2.

Using the inverse sine function, we can find the principal value of 2θ which is π/6 radians. This gives us one solution for θ: π/12.

Since sin function is periodic with period of 2π, we can find additional solutions by adding integer multiples of π to the principal value. Therefore, the solutions for θ are π/12 + kπ, where k is any integer.

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Answer:

f(13)=2x-1, x=7

Step-by-step explanation:

7*2=14, 14-1=13

vise versa

f(13)=2(7)-1

Answer: 25

Step-by-step explanation: Easy. Just substitute 13 for x into the equation. f(13)=2(13)-1

= 26-1

= 25

Answer:

7) d)

standard error of the mean of one sample of 'n' observation = 0.20

8) a)

The margin of Error = 0.392

9) d

The 95% of confidence intervals are (8.61 , 9.39)

Step-by-step explanation:

7)

solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

The standard error of the mean of one sample of 'n' observation is

Standard error (SE)

                               = [tex]\frac{S.D}{\sqrt{n} }[/tex]  

                               = σ / √n

                               = [tex]\frac{1.8}{\sqrt{81} } =0.2[/tex]

standard error of the mean of one sample of 'n' observation = 0.20

8)

Solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

Given the probability is 0.95

The z- score = 1.96 at 0.05 level of significance.

The margin of Error   =  [tex]\frac{z_{0.95} S.D}{\sqrt{n} }[/tex]

                                   = [tex]\frac{1.96 (S.D)}{\sqrt{n} }[/tex]

                                   = [tex]\frac{1.96 (1.8)}{\sqrt{81} }[/tex]

                                   = 0.392

The margin of Error = 0.392

9)

Solution:-

The 95% of confidence intervals are

[tex](x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96\frac{S.D}{\sqrt{n} } )[/tex]

[tex](9 - 1.96\frac{1.8}{\sqrt{81} } , 9+ 1.96\frac{1.8}{\sqrt{81} } )[/tex]

(9 - 0.392 , (9 + 0.392)

(8.609 , 9.392)

The 95% of confidence intervals are (8.61 , 9.39)

Answer:

[tex] \bigg(\frac{2}{5} \bigg)^{4} [/tex]

Step-by-step explanation:

[tex] \frac{2}{5} \times \frac{2}{5} \times \frac{2}{5} \times \frac{2}{5} = \bigg(\frac{2}{5} \bigg)^{4} \\ [/tex]

Answer:

185

Step-by-step explanation:

cause if you add up 80+65+40 right?

To find the overall probability of leaving on-time, we consider the probabilities of each airline and their respective on-time rates. The weighted average of the on-time rates is calculated using the percentages of scheduled flights for each airline. Summing up the weighted contributions gives us the overall probability of leaving on-time.

To find the overall probability of leaving on-time, we need to consider the probabilities of each airline and their respective on-time rates. First, we calculate the weighted average of the on-time rates using the percentages of scheduled flights for each airline.

Airline A contributes 50% of the flights with an 80% on-time rate, so its weighted contribution is 0.5 x 0.8 = 0.4.

Airline B contributes 30% of the flights with a 65% on-time rate, so its weighted contribution is 0.3 x 0.65 = 0.195.

Airline C contributes 20% of the flights with a 40% on-time rate, so its weighted contribution is 0.2 x 0.4 = 0.08.

To find the overall probability, we sum up the weighted contributions: 0.4 + 0.195 + 0.08 = 0.675.

Therefore, the overall probability of leaving on-time is 0.675 or 67.5%.

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Answer:

The proportion of slots which meet these specifications is 0.16634 or 16.63%.

Step-by-step explanation:

We are given that the width in inches of slots cut by a milling machine follows approximately the N(0.72,0.0012) distribution.

Also, the specifications allow slot widths between 0.71975 and 0.72025.

Let X = width in inches of slots cut by a milling machine

The z-score probability distribution for normal distribution is given by;

                           Z = [tex]\frac{ X-\mu}{{\sigma}} }} }[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean width = 0.72

            [tex]\sigma[/tex] = standard deviation = 0.0012

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the specifications allow slot widths between 0.71975 and 0.72025 is given by = P(0.71975 < X < 0.72025)

      P(0.71975 < X < 0.72025)  = P(X < 0.72025) - P(X [tex]\leq[/tex] 0.71975)

     P(X < 0.72025) = P( [tex]\frac{ X-\mu}{{\sigma}} }} }[/tex] < [tex]\frac{ 0.72025-0.72}{{0.0012}} }} }[/tex] ) = P(Z < 0.21) = 0.58317

     P(X [tex]\leq[/tex] 0.71975) = P( [tex]\frac{ X-\mu}{{\sigma}} }} }[/tex] [tex]\leq[/tex] [tex]\frac{ 0.71975-0.72}{{0.0012}} }} }[/tex] ) = P(Z [tex]\leq[/tex] -0.21) = 1 - P(Z < 0.21)

                                                                    = 1 - 0.58317 = 0.41683

So, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.21 in the z table which has an area of 0.58317.

Therefore, P(0.71975 < X < 0.72025)  = 0.58317 - 0.41683 = 0.16634

Hence, the proportion of slots who meet these specifications is 16.63%.

Given:

Given that the rectangular prism with length l = 2..6 yards.

The width of the rectangular prism is 3.5 yards.

The height of the prism is h yards.

We need to determine the simplified expression for the surface area of the rectangular prism.

We also need to determine the surface area of the prism if the height is 4 yards.

Expression for the surface area of the prism:

The surface area of the rectangular prism can be determined using the formula,

[tex]SA=2(lw+wh+lh)[/tex]

Substituting l = 2.6, w = 3.5, h = h, we get;

[tex]SA=2[(2.6)(3.5)+(3.5)h+(2.6)h][/tex]

[tex]SA=2(9.1+3.5h+2.6h)[/tex]

[tex]SA=2(9.1+6.1 h)[/tex]

[tex]SA=18.2+12.2h[/tex]

Thus, the simplified expression for the surface area of the prism is [tex]SA=18.2+12.2h[/tex]

Surface area of the prism:

Now, we shall determine the surface area of the prism if the height is 4 yards.

Substituting the value h = 4 in the expression [tex]SA=18.2+12.2h[/tex], we have;

[tex]SA=18.2+12.2(4)[/tex]

[tex]SA=18.2+48.8[/tex]

[tex]SA=67 \ yd^2[/tex]

Thus, the surface area of the rectangular prism is 67 square yards.

Answer:

d) A two-sample z-interval for a difference in sample proportions

Step-by-step explanation:

Explanation:-

Given data a random sample of 40 engines of one design, 14 failed to ignite as a result of fuel system error.

First sample proportion    [tex]p_{1} = \frac{14}{40} = 0.35[/tex]

Given data random sample of 30 engines of a second design, 9 failed to ignite as a result of fuel system error.

second sample proportion   [tex]p_{2} = \frac{9}{30} = 0.30[/tex]

Null hypothesis: H₀: Assume that there is no significant between the two designs

H₀: p₁ = p₂

Alternative Hypothesis: H₁:

H₁: p₁ ≠ p₂

The test statistic

                          [tex]Z = \frac{p_{1}-p_{2} }{\sqrt{p q(\frac{1}{n_{1} } + \frac{1}{n_{2} } } )}[/tex]

where [tex]p = \frac{n_{1} p_{1}+n_{2} p_{2} }{n_{1} +n_{2} }[/tex]

          q =1-p

Answer:

E

Step-by-step explanation:

A two-sample z-interval for a difference in population proportions

Answer:

3 4/7 = 25/7

Step-by-step explanation:

multiply the whole number(3) by the denominator(7).

Then you have 21.

Add the product of the last 2 numbers to the numerator(4)

21 + 4 = 25

Note: (denominator x whole number) + numerator

We are required to convert 3 and 4/7 to an improper fraction.

The fraction 3 and 4/7 to an improper fraction is 25 / 7

Given:

3 4/7

= {7(3) + 4} / 7

= (21 + 4) / 7

= 25 / 7

Therefore, the fraction 3 and 4/7 to an improper fraction is 25 / 7

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The total lifetime value for 5% of packages to exceed is approximately 103.94 hours.

To solve this problem, we first need to find the distribution of the total lifetime of all nine batteries in a package. Since each battery's lifetime follows a normal distribution with a mean of 11 hours and a standard deviation of 1 hour, the total lifetime of all nine batteries will also follow a normal distribution.

The mean of the total lifetime of all nine batteries is [tex]\( 9 \times 11 = 99 \)[/tex] hours.

The standard deviation of the total lifetime of all nine batteries is [tex]\( \sqrt{9} \times 1 = 3 \)[/tex] hours.

Now, we need to find the value such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages. This is equivalent to finding the 95th percentile of the distribution of the total lifetime.

We'll use the Z-score formula to find the Z-score corresponding to the 95th percentile, and then use that Z-score to find the corresponding value in terms of hours.

The Z-score corresponding to the 95th percentile is approximately 1.645 (you can find this value from standard normal distribution tables or calculators).

Now, we'll use the formula:

[tex]\[ \text{Value} = \text{Mean} + (\text{Z-score} \times \text{Standard deviation}) \][/tex]

[tex]\[ \text{Value} = 99 + (1.645 \times 3) \][/tex]

[tex]\[ \text{Value} = 99 + 4.935 \][/tex]

[tex]\[ \text{Value} \approx 103.94 \][/tex]

So, the total lifetime value such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages is approximately 103.94 hours.

Complete Question:

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. There are nine batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (Round your answer to two decimal places.)

______ hours

Answer:

You did not include your graph, therefore I will explain in the step-by-step everything you should need to know.

Step-by-step explanation:

The standard slope formula is y=mx + b, where m = slope (rise/run) and b = the y intercept.

Using this knowledge, we know that this line has a slope of 5, or 5/1, which means that you will go up 5 and 1 to the right. Since there is no + b part, that means b = 0, so the line begins at 0,0. So, using this information, possible points include 0,0, 5,1, 10,2, 15,3, etc.

Answer:

Renaming a fraction

Step-by-step explanation:

Answer:

1. renaming a fraction

2. de-c

Step-by-step explanation:

   Andy Beth Tri Dale Ella.

To solve the problem using Polya's four-step method, assign variables to represent the runners, set up a system of equations, and solve for the unknowns.

To solve this problem using Polya's four-step method, we need to identify the given information and the unknowns. Let's assign variables to represent the runners: Andy (A), Beth (B), Dale (D), Ella (E), and Tri (T). From the given information, we know that Dale finished 5 seconds before Tri (D - T = 5), Tri finished 7 seconds after Beth (T - B = 7), Beth finished 7 seconds after Ella (B - E = 7), and Ella finished 4 seconds before Andy (E - A = 4). Now, we can set up a system of equations to solve for the order of the runners.
From the equations, we can solve for the values of the variables. Plugging the values back into the original equations, we find that the order in which the runners finished the race is Andy, Ella, Beth, Tri, and Dale.

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Answer:

17 adult tickets

23 children's tickets

Step-by-step explanation:

Adult tickets 17x7=119

Child tickets 23x5=115                     23 is 6 more than 17.

119+115=234

Given Information:

Cost price of lettuce = $10

Selling price of lettuce = $21

Salvage value = $2

Average demand of lettuce = μ = 262 boxes

Standard deviation of lettuce = σ = 43 boxes

Required Information:

How many boxes of lettuce should the supermarket purchase = ?

Answer:

n = 271 boxes

Step-by-step explanation:

The required number of lettuce boxes that supermarket should purchase is given by

n = μ + (z-score)σ

Where μ is the average demand of lettuce boxes, σ is the standard deviation, and z-score can be calculated by

p = C_us/(C_us + C_os)

Where the cost of under stocking is given by

C_us = Selling price of lettuce - Cost price of lettuce

C_us = $21 - $10

C_us = $11

The cost of over stocking is given by

C_os = Cost price of lettuce - Salvage value

C_os = $10 - $2

C_os = $8

p = C_us/(C_us + C_os)

p = 11/(11 + 8)

p ≈ 58%

The z-score corresponding to 58% is 0.202

n = 262 + (0.202)43

n = 270.68

n = 271 boxes

Therefore, the supermarket should purchase 271 boxes of lettuce tomorrow.

Answer:

  $28

Step-by-step explanation:

The markup was 0.40×$20 = $8, so the new price is ...

 $20 +8 = $28

Answer:

The minimum score required for an A grade is 89.8.

Step-by-step explanation:

We are given that a psychology professor assigns letter grades on a test according to the following scheme. A : Top 7% of scores. B : Scores below the top 7% and above the bottom 64%. C : Scores below the top 36% and above the bottom 25%. D : Scores below the top 75% and above the bottom 6%. F : Bottom 6% of scores.

Scores on the test are normally distributed with a mean of 78.4 and a standard deviation of 7.6.

Let X = Scores on the test

SO, X ~ Normal([tex]\mu=78.4,\sigma^{2} =7.6^{2}[/tex])

The z-score probability distribution for normal distribution is given by;

                              Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = mean time = 78.4

            [tex]\sigma[/tex] = standard deviation = 7.6

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the minimum score required for an A grade so that it represents Top 7% of scores is given by;

      P(X [tex]\geq[/tex] x) = 0.07   {where x is the required minimum score

      P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{x-78.4}{7.6}[/tex] ) = 0.07

       P(Z [tex]\geq[/tex] [tex]\frac{x-78.4}{7.6}[/tex] ) = 0.07

So, the critical value of x in the z table which represents the top 7% of the area is given as 1.4996, that is;

                        [tex]\frac{x-78.4}{7.6} =1.4996[/tex]

                        [tex]{x-78.4}{} =1.4996\times 7.6[/tex]

                        [tex]x[/tex]  = 78.4 + 11.39696 = 89.8 or 90

Hence, the minimum score required for an A grade is 89.8.

Answer:

[tex]z=-2.33<\frac{a-8.9}{1.9}[/tex]

And if we solve for a we got

[tex]a=8.9 -2.33*1.9=4.47[/tex]

And the best answer for this case would be:

A) 4.47 lb

Step-by-step explanation:

Let X the random variable that represent the weights of juvenile turkeys, and for this case we know the distribution for X is given by:

[tex]X \sim N(8.9,1.9)[/tex]  

Where [tex]\mu=8.9[/tex] and [tex]\sigma=1.9[/tex]

The z score formula very useful for this case is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.99[/tex]   (a)

[tex]P(X<a)=0.01[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. On this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.99

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.01[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.01[/tex]

So we have this relation

[tex]z=-2.33<\frac{a-8.9}{1.9}[/tex]

And if we solve for a we got

[tex]a=8.9 -2.33*1.9=4.47[/tex]

And the best answer for this case would be:

A) 4.47 lb

Answer:

Anna33 diane11

Step-by-step explanation:

11×3=33

33+11=44