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3 years ago

16. Suppose Eva has 12 3/4 pounds of flour. The day's baking will require triple that amount.

Mathematics

1 answer:

egoroff_w [7]3 years ago

3 0

Answer:

Option D

Step-by-step explanation:

Given the following question:

$12 \frac{3}{4} \times3$

In order to find the answer, we have to convert the mixed number and then multiply by three, since how Eva needs triple the amount for today's baking.

$12 \frac{3}{4} \times3$
$12\frac{3}{4} =4\times12=48+3=51=\frac{51}{4}$
$3=\frac{3}{1}$
$\frac{51}{4} \times\frac{3}{1}$
$51\times3=153$
$4\times1=4$
$=\frac{153}{4}$

<u>Convert the improper fraction back into a mixed number:</u>

$\frac{153}{4}$
$\frac{153}{4} =153\div4=38\frac{1}{4}$
$=38\frac{1}{4}$

Which means Eva needs "38 1/4 pounds of flour" or option D for today's baking.

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2 years ago

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Answer:

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that $\mu = 118, \sigma = 8$

Sample of 16 shells

This means that $n = 16, s = \frac{8}{\sqrt{16}} = 2$

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{120 - 118}{2}$

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$Z = 1$ has a pvalue of 0.841

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$Z = \frac{X - \mu}{s}$

$Z = \frac{116 - 118}{2}$

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$Z = -1$ has a pvalue of 0.159

0.841 - 0.159 = 0.682

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

3 0

3 years ago