MATHEMATICS COLLEGE

8.) Out of 600 employees, only 270 are happy with their pay. What percent is this? (Please show your work.) *

Answers

Answer 1

Answer:

For this case we can raise a rule of three:

600 employees -------------> 100%

270 employees -------------> x

Where the variable "x" represents the percentage of employees who are satisfied with their salary. So, we have:

$x = \frac {270 * 100} {600}\nx = 45$

Thus, 45% of employees are satisfied with their salary.

Answer:

45%

Answer 2

Answer:

ANSWER

$45\%$

EXPLANATION

The total number of employees is 600.

The number of employees who are happy with their pay is 270.

The percentage of employees who are happy with their pay is the number who are happy with their pay divided by total number of employees times 100%

$(270)/(600) * 100\%$

This simplifies to

$45\%$

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Suppose that you are examining the difference in ages between brides and grooms. You are interested in conducting a significance test to examine to address the theory that the bride is younger than the groom in more than half of all marriages. A sample of 100 couples were observed, for which 67 had a bride younger than the groom.whats is the parameter? .
Step 2: The null hypothesis and alternative hypothesis?
Step 5: whats the p-value is:

Answers

Answer:

Parameter = 0.5

Null hypothesis : H0 : P0 = 0.5

Alternative hypothesis ; H0 : P0 > 0.5

Pvalue = 0.99966

Step-by-step explanation:

The parameter defines a statistical value or calculation which is derived from the population.

The parameter in this scenario is the population proportion, P0 which is 0.5

The scenario above describes a scenario to test the difference in population.

The null hypothesis, that bride and groom are of the same age ;

H0 : P0 = 0.5

The alternative hypothesis ; the bride is younger Than the groom in more than half of the population.

H1 : p0 > 0.5

To obtain the Pvalue :

Test statistic : (phat - P0) ÷ √(p0(1 -p0) / n)

Phat = x/n

x = 67 ; sample size, n = 100

Phat = x / n = 67/100 = 0.67

P0 = 1 - 0.5 = 0.5

Tstatistic = (0.67-0.50) ÷ √(0.5(0.5) / 100)

Tstatiatic = 0.17 ÷ 0.05

Tstatistic = 3.4

P-value : p(Z < 3.4) = 0.99966 (Z probability calculator).

In a study of factors affecting whether soldiers decide to reenlist, 320 subjects were measured for an index of satisfaction. The sample mean is 28.8 and the sample standard deviation is 7.3. Use the given sample data to construct the 98 percent confidence interval for the population mean.

Answers

Answer:

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1-0.98)/(2) = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.01 = 0.99$ , so $z = 2.325$

Now, find M as such

$M = z*(\sigma)/(√(n))$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.325(7.3)/(√(320)) = 0.9488$

The lower end of the interval is the mean subtracted by M. So it is 28.8 - 0.9488 = 27.85 subjects.

The upper end of the interval is the mean added to M. So it is 28.8 + 0.9488 = 29.75 subjects.

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.

Which value of x makes this equation true -5(x - 20) = 35

Answers

The value of $x$ that makes the equation true is $$x = 13$.$

To find the value of $x$ that makes the equation $-5(x - 20) = 35$ true, you need to simplify the equation and solve for $x$. Let's break down the steps: