Anna learns in her science class that the human eye blinks an average of 4,200,000 times a year. What is this number expressed in scientific notation?

8 0

3 years ago

Solve for x. x/3≤—6 1.x≤—2 2.x≤—18 3.x≥—2 4.x≥—18

RSB [31]

Answer:

$x \leqslant - 18$

Step-by-step explanation:

Multiply both sides by 3

$x \leqslant - 18$

6 0

3 years ago

2 + 2 = ? please help!

dybincka [34]

Answer:

Hope that this helps!

7 0

3 years ago

Read 2 more answers

A sales and marketing management magazine conducted a survey on salespeople cheating on their

Alenkinab [10]

Answer:

The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).

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 $1 - \frac{\alpha}{2}$ .

In the survey on 200 managers, 58% of the managers have caught salespeople cheating on an expense report.

This means that $n = 200, \pi = 0.58$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 - 1.96\sqrt{\frac{0.58*0.42}{200}} = 0.5116$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 + 1.96\sqrt{\frac{0.58*0.42}{200}} = 0.6484$

The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).

7 0

3 years ago