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3 years ago

Simplify the expression. (7−2)24+34−10 13 35 53 3

Mathematics

1 answer:

Mashcka [7]3 years ago

8 0

Answer:

144

Step-by-step explanation:

Subtract 2 from 7 to get 5.

5×24+34−10=144

Multiply 5 and 24 to get 120.

120+34−10=144

Add 120 and 34 to get 154.

154−10=144

Subtract 10 from 154 to get 144.

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Can u put 0.5,0.41 and 3/5 in least to greatest

baherus [9]

0.41, 0.5, 3/5 is the order

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3 years ago

An environmentalist wants to find out the fraction of oil tankers that have spills each month. Step 2 of 2 : Suppose a sample of

rewona [7]

Answer:

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

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 333 tankers is drawn. Of these ships, 257 did not have spills.

333 - 257 = 76 have spills.

This means that $n = 333, \pi = \frac{76}{333} = 0.228$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 - 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.199$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 + 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.257$

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

6 0

3 years ago

Suppose that time spent on hold per call with customer service at a large telecom company is normally distributed with a mean µ

lana66690 [7]

Answer:

0.3108 is the probability that the sample mean is between 7.8 and 8.2 minutes.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8 minutes

Standard Deviation, σ = 2.5 minutes

Sample size, n = 25

We are given that the distribution of time spent is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{2.5}{\sqrt{25}} = 0.5$

P(sample mean is between 7.8 and 8.2 minutes)

$P(7.8 \leq x \leq 8.2)\\\\ = P(\displaystyle\frac{7.8 - 8}{0.5} \leq z \leq \displaystyle\frac{8.2-8}{0.5})\\\\ = P(-0.4 \leq z \leq 0.4})\\\\= P(z < 0.4) - P(z < -0.4)\\\\= 0.6554 -0.3446= 0.3108$