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3 years ago

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Mathematics

1 answer:

alukav5142 [94]3 years ago

8 0

Answer:

D. 70

Step-by-step explanation:

if vx is bisector then 3z - 4 = z + 6 add like terms

2z = 10;and z = 5

To find the perimeter we add all side lengths

VU + UW + WV = 5z - 1 + 5z - 1 + 3z - 4 + z + 6 = 14z

we know z = 5 so 14 × 5 = 70

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Work out the value of 7^2-3^3

kow [346]

Answer:

Step-by-step explanation:

7^2=49

3^3=27

49-27=22

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3 years ago

Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample

fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

37% of the company's orders come from first-time customers.

This means that $p = 0.37$

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that $n = 225$

Mean and standard deviation:

$\mu = p = 0.37$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322$

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

$Z = \frac{0.38 - 0.37}{0.0322}$

$Z = 0.31$

$Z = 0.31$ has a pvalue of 0.6217

X = 0.26

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

$Z = \frac{0.26 - 0.37}{0.0322}$

$Z = -3.42$

$Z = -3.42$ has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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3 years ago

The mean cholesterol level in children is 175 mg/dl with standard deviation 35 mg/dl. Assume this level varies from child to chi

dsp73

Answer: Read the following,

Step-by-step explanation:

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3 years ago

A company did a quality check on all the packs of nuts it manufactured. Each pack of nuts is targeted to weigh 18.25 oz. A pack

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Answer:

$x < 17.89$ or $x > 18.61$ , since $\left | x-18.25 \right |>0.36$

Step-by-step explanation:

According to the question the weight is rejected if it is more than 0.36 oz, the weight should not be beyond 0.36 oz that is from the target weight of 18.25.

We can say that, the difference between the actual weight and the target weight is greater than 0.36 oz.

Lets say the actual weight is 'x'. So,

$x-18.25>0.36$