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3 years ago

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A document you fill out when applying for a job. It asks for information about your work history, education and personal referen

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kow [346]3 years ago

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the correct awners is a application

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Shtirlitz [24]

x(x + 9-x) = 6^2

9x = 36

x = 4

Answer

c. 4

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3 years ago

ILL GIVE BRAINLIEST!! PLS HELP!!! Ten students are taking both algebra and drafting. There are 24 students taking algebra. There

Minchanka [31]

Answer:

15 students

Step-by-step explanation:

from the algebra = 24 - 10 = 14

from the draft = 11 - 10 = 1

14 + 1 = 15

the total students that are taking algebra or drafting but not both is 15 students

4 0

3 years ago

Read 2 more answers

Can a triangle have side lengths 4, 4, 7? <br> 1.no <br> 2. Yes <br> 3. not enough information

Marina CMI [18]

Yes! The lengths of each side must be less than the sum of the other two lengths. It looks like this:
4+4>7
7+4>4
4+7>4

8 0

3 years ago

A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

The confidence interval is given by this formula:

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

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

8 0

3 years ago

WILL GIVE BRAINLIEST

Karolina [17]

1. B(15 to 42)
2. B(8 to 20)
3. C(2:1)

5 0

4 years ago

Read 2 more answers