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skad [1K]
3 years ago
13

Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights

of oranges are also normally distributed with a mean of 131 grams and a standard deviation of 20 grams. Amy has an apple that weighs 90 grams and an orange that weighs 155 grams.
Required:
a. Find the probability a randomly chosen apple exceeds 100 g in weight.
b. What weight do 80% of the apples exceed?
Mathematics
1 answer:
user100 [1]3 years ago
5 0

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

Z = \frac{X - \mu}{\sigma}

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

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Pyramid A is a square pyramid with a base side length of 12 inches and a height of 8 inches. Pyramid B is a square pyramid with
yan [13]

Answer:

<em>The volume of pyramid B is 64 times the volume of pyramid A.</em>

<em></em>

Step-by-step explanation:

Given:

Two square pyramids A and B.

Side Length of A, s_A = 12 inches

Height of A, h_A = 8 inches

Side Length of B, s_B = 48 inches

Height of B, h_B = 32 inches

To find:

How many times bigger is the volume of pyramid B than pyramid A?

OR

V_B is how many times bigger than V_A ?

Solution:

First of all, let us have a look at the formula for volume of a pyramid:

V=\dfrac{1}{3} \times \text{Area of Base} \times \text{Height}

Here, base is square, so:

V=\dfrac{1}{3} \times s^2 \times h

Volume of pyramid A:

V_A=\dfrac{1}{3} \times s_A^2 \times h_A

\Rightarrow V_A=\dfrac{1}{3} \times 12^2 \times 8 = 384\ inch^3

Volume of pyramid B:

V_B=\dfrac{1}{3} \times s_B^2 \times h_B

\Rightarrow V_B=\dfrac{1}{3} \times 48^2 \times 32 \\\Rightarrow V_B=\dfrac{1}{3} \times 12^2 \times 8 \times 4^2 \times 4\\\Rightarrow V_B=\dfrac{1}{3} \times 12^2 \times 8 \times 64\\\Rightarrow V_B= 24576\ inch^3 = 384\times 64\ inch^3\\\Rightarrow V_B = V_A\times 64\ inch^3

<em>The volume of pyramid B is 64 times the volume of pyramid A.</em>

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Lelechka [254]

Answer:

On a coordinate plane, a trapezoid has points (0, 0), (0.5, 2), (1, 2), (1.5, 0).

Step-by-step explanation:

The given trapezoid has vertices at

(0, 0), (1, 4), (2, 4), (3, 0).

We want to choose from the given options the trapezoid with a scale factor between 0 and 1.

If a scale factor of a dilation is between 0 and 1, the size of the image trapezoid will be smaller than the preimage.

When we analyze the coordinates carefully, the answer is, a trapezoid has points (0, 0), (0.5, 2), (1, 2), (1.5, 0).

The reason is that, when we multiply the coordinates of (0, 0), (1, 4), (2, 4), (3, 0) by 0.5, we obtain (0, 0), (0.5, 2), (1, 2), (1.5, 0).

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