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Mathematics

2 answers:

Wewaii [24]3 years ago

8 0

<em> $answer =50.24 {yd}^{2} \\ solution \\ diameter = 8 \: yards \\ radius = \frac{8}{2} = 4 \: yards \\ area \: of \: circle = \pi \: {r}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: 3.14 \times {4}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 3.14 \times 16 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 50.24 \: {yd}^{2} \\ hope \: it \: helps \\ good \: luck \: on \: your \: assignment$ </em>

7nadin3 [17]3 years ago

3 0

Answer:

$50.24 yd^2$

Step-by-step explanation:

Find the radius from the diameter given.

$r=d/2$

$r=8/2$

$r=4$

Use the area of circle formula.

$\pi r^2$

$3.14 \times 4^2$

$16 \times 3.14$

$=50.24$

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

user100 [1]

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 $\mu = 85, \sigma = 8$