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3 years ago

(MULTIPLE CHOICE QUESTION)

Mathematics

2 answers:

Dmitrij [34]3 years ago

6 0

Answer:

I believe the last is does help while the rest don't help.

Step-by-step explanation:

The first only marks O and M

The second uses O and M and only finds 2 unrelated points.

The third uses O and M to find the other 3 points

Flura [38]3 years ago

4 0

Answer: i think it’s the last one

Step-by-step explanation:

You might be interested in

Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin

NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Confidence interval

Assuming that $\bar X =5.5$ and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=1086-1=1085$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that $t_{\alpha/2}=2.33$

Now we have everything in order to replace into formula (1):

$5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139$

$5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861$

So on this case the 95% confidence interval would be given by (5.139;5.861) We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

5 0

3 years ago

Which type of statement switches and negates the hypothesis and the conclusion of the original conditional statement?

valentinak56 [21]

Inverse

Inverse - is a conditional statement that NEGATES both the hypothesis and the conclusion of the original conditional statement.

8 0

3 years ago

I need help with this ^^ <br> It’s about choosing a graph to fit a narrative

jenyasd209 [6]

1. The fourth graph: D
2. The first graph: A

3 0

3 years ago

After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only five women among the last

Georgia [21]

Answer:

The probability that 5 or fewer women are hired, assuming no gender discrimination, is 0.0317; we can use this result to support her charge of gender discrimination.

Step-by-step explanation:

If we are assuming that the women and the men are equally qualified, then the probability for each employee that is hired the probability for it to be a women should be 1/2. Note that the fact that more men that women are hired in a sample might not be disctrimination: for example, if 2 men are hired out of 2 employees, that can happen with probability 1/4, so it is quite common. In order to support her charge for gender discrimination, we need at least a probability less that 0.05 that 5 (or less) women are hired out of 19 employees.

Since each configuration is equally probable, we will count the total amount of possible cases that 5 or less women are hired, and dividide it by the total amount of cases, 2¹⁹.

0 women hired: one possible case: every employee is male
1 women hired: 19 possible cases
2 women hired: ${19 \choose 2} = 171$ possible cases
3 women hired: ${19 \choose 3} = 969$ possible cases
4 women hired: ${19 \choose 4} = 3876$ possible cases
5 women hired: ${19 \choose 5} = 11628$ possible cases

Thus, there are a total of 11628+3876+969+171 = 16644 possible cases out of 2¹⁹ ones. All of them with seemingly equal probability. As a consequence, the probability of 5 or less women to be hired out of 19 employees, assuming that the probability to hire 1 is 1/2, is

16644/2¹⁹ = 0.0317 < 0.05

The probability that 5 or fewer women are hired, assuming no gender discrimination, is 0.0317. Since the probability is so low, we can conclude that for the employer, a woman equally qualified as a man is less likely to be hired, therefore, we can support her charge of gender discrimination.

6 0

3 years ago

2. (01.07 MC)

Elena-2011 [213]

Part A: subtract 6 from both sides
Divide by -3 on both sides
X=-3
Part B: add like terms (-2K-3k)
Add 12 to both sides
Add 5k to both sides
Divide by 5 on both sides
K=3
Part C: distribute 6 into the parentheses
Add like terms together (36-5)
Subtract 1 from both sides
Subtract 36v from both sides
Divide by -30 on both sides
-1=v

7 0

3 years ago