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2 years ago

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Anthony currently earns $11.75 per hour. • He will get a $0.25 raise after 6 months. • He will get a 2.5% raise after an additio

nal 6 months. After Anthony gets both raises, what will his pay be for 38 hours of work?

1 answer:

Usimov [2.4K]2 years ago

4 0

Answer:

You need to give more information that that...

Step-by-step explanation:

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H(x)=2x: Find (7)<br><br>please help

erma4kov [3.2K]

Answer:

2(7) = 14

Step-by-step explanation:

When there is a number in the brackets, we want to input that into our function and get the output. In other words, we simply substitute 7 for x

2(7) = 14

3 0

2 years ago

Charlotte reads 8 1/3 pages of a book in 10 minutes. What is her average reading rate in pages per minute?

sukhopar [10]

$8\frac{1}{3} = \frac{25}{3}$

25/3 pages of a book in 10 minutes

Multiply both numbers by the reciprocal of 10, which is 1/10

$\frac{25}{3} \times\frac{1}{10} = \frac{25}{30}=\frac{5}{6}$

$10\times\frac{1}{10}= 1$

So, Charlotte reads 5/6 pages of a book in 1 minute.

8 0

3 years ago

Use the Central Limit Theorem to find a mean given a probability Question A video game company sells an average of 132 games a m

aev [14]

Answer:

The number of games must a store sell in order to be eligible for a reward is 135.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of video games sold in a month by the sores.

The random variable <em>X</em> has a mean of, <em>μ</em> = 132 and a standard deviation of, <em>σ</em> = 9.

It is provided that the company is looking to reward stores that are selling in the top 7%.

That is, $P (\bar X > \bar x) = 0.07$ .

The <em>z</em>-score related to this probability is, <em>z</em> = 1.48.

Compute the number of games must a store sell in order to be eligible for a reward as follows:

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}$

$\bar x=\mu+z\cdot \sigma/\sqrt{n}$

$=132+1.48\times (9/\sqrt{36})\\\\=132+2.22\\\\=134.22\\\\\approx 135$

Thus, the number of games must a store sell in order to be eligible for a reward is 135.

5 0

3 years ago

What does reflects most nearly mean

Gala2k [10]

Either to mirror, or to copy.

5 0

2 years ago

Read 2 more answers

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :

const2013 [10]

Answer:

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that $n = 1067, \pi = \frac{74}{1067} = 0.069$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079$

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

7 0

2 years ago