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densk [106]
3 years ago
15

estimate the answer by rounding first number to nearest tenth and second number to nearest hundred. 0.65 × 1754. have to type an

integer or decimal​
Mathematics
1 answer:
Levart [38]3 years ago
7 0

Answer:

1260

Step-by-step explanation:

Round 0.65 to 0.7 and round 1754 to 1800. To get 1260, I recommend an easy way to do this. I would take 18 and multiply it by 7, giving us 126. Because it was supposed to be 0.7, not 7, we multiply by 10 because 7/10 is 0.7. This gives us 126 times 10 which equals 1260.

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For a large supermarket chain in a particularâ state, aâ women's group claimed that female employees were passed over for manage
PolarNik [594]

Answer:

1) As the P-value (0.097) is greater than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the women's claim that female employees were passed over for management training in favor of their male colleagues.

2) The 95% confidence interval for the population proportion is (0.173, 0.427).

The populations proportion (π=0.4) is included in this confidence interval, so the claim has no enough evidence to be supported.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that female employees were passed over for management training in favor of their male colleagues.

We use the women's part for sample and population proportions.

Then, the null and alternative hypothesis are:

H_0: \pi=0.4\\\\H_a:\pi

The significance level is 0.05.

The sample has a size n=50 persons.

The sample proportion is p=0.3.

p=X/n=15/50=0.3

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.4*0.6}{50}}\\\\\\ \sigma_p=\sqrt{0.0048}=0.069

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.3-0.4+0.5/50}{0.069}=\dfrac{-0.09}{0.069}=-1.299

This test is a left-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z

As the P-value (0.097) is greater than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that female employees were passed over for management training in favor of their male colleagues.

b) We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.3.

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.3*0.7}{50}}\\\\\\ \sigma_p=\sqrt{0.0042}=0.0648

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_p=1.96 \cdot 0.0648=0.127

Then, the lower and upper bounds of the confidence interval are:

LL=p-z \cdot \sigma_p = 0.3-0.127=0.173\\\\UL=p+z \cdot \sigma_p = 0.3+0.127=0.427

The 95% confidence interval for the population proportion is (0.173, 0.427).

The populations proportion (π=0.4) is included in this confidence interval, so the claim has no enough evidence to be supported.

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zaharov [31]

Your answer is 2x+28

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