Is 2/13a repeating decimal

Jason wants to purchase a new pair of shoes. He has $70 he is able to spend. A popular new pair of shoes are sold at the shoe st

Ghella [55]

Answer:

$72.76 Us Dollars

Step-by-step explanation:

1st I found out what 20% off of $85 was and it was $68 US Dollars then I had to add 7% of $68 US Dollars and it came out to $72.76

Please Mark Brainliest!!!!

8 0

3 years ago

What multiplication problem is being solved?

gulaghasi [49]

Answer is -15 5×-3 = -15.

7 0

4 years ago

Cody wants to join the army for a career and he also wants a wife and kid. If he gets paid $25000 a year will he be able to affo

RSB [31]

Yes, if that is the expenses he has to pay for all three of them. All you have to do is add up the all his expenses and multiple by 12

6 0

3 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, 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.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

6 0

3 years ago

A rectangular box has three of its faces on the coordinate planes and one vertex in the first octant on the paraboloid z=100−x2−

LuckyWell [14K]

The volume as a function of the location of that vertex is

... v(x, y, z) = x·y·z = x·y·(100-x²-y²)

This function is symmetrical in x and y, so will be a maximum when x=y. That is, you wish to maximize the function

... v(x) = x²(100 -2x²) = 2x²(50-x²)

This is a quadratic in x² that has zeros at x²=0 and x²=50. It will have a maximum halfway between those zeros, at x²=25. That maximum volume is

... v(5) = 2·25·(50-25) = 1250

The maximum volume of the box is 1250 cubic units.

7 0

3 years ago