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2 years ago

An employee earns $255 for 15 hours of work. Assuming he is paid by the hour, how much will he earn in 8 hours?

2 answers:

7 0

First you do $255/15= 17 dollars per hour. Then you have to multiply 17*8= 136. So he would get $136 in 8 hrs.

Hope this helps!

Talja [164]2 years ago

4 0

First divide 255 from 15 hours to get the amount he gets payed per hour

255/15 = 17 dollars per hour

Then multiply 17 to 8 hours to get the total for 8 hours

17 x 8 = 136

He makes $136 in 8 hours

hope this helps

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At a movie theatre adult tickets cost $10 and child ticjets cost $6. Which equation can be used to find s, the number of dollars

timurjin [86]

Answer:

s = 10k + 30

Step-by-step explanation:

Cost of Adults ticket = $10 per adult

Cost of Ticket = $6 per child

Number of adults = k

Number of Children = 5

s = Total cost

Total cost = (Cost per adult * number of adults) + (Cost per child * Number of children)

s = ($10 * k) + ($6 * 5)

s = $10k + $30

s = 10k + 30

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Answer:

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Answer:

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Step-by-step explanation:

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Grace [21]

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Read 2 more answers

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a standard deviation of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

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

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

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3 years ago