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Umnica [9.8K]
3 years ago
5

Could you please name the place value to which each number was rounded

Mathematics
1 answer:
cluponka [151]3 years ago
7 0

Answer:

QUESTION NOT CLEARLY STATED

Step-by-step explanation:

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PLEASE HELP ASAP!!!
Eva8 [605]
I'm not 100% sure that this is correct but it's a 50 50

The first one is none of these

Second one is alternate interior (I am 100% sure on this one)

Third one is corresponding

Fourth one is alternate exterior (also 100% sure)

Fifth one is also corresponding

I tried my best

7 0
4 years ago
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Yapın.............. Bir sürpriz.. yardımcı olur musunuz ​
borishaifa [10]

Answer:

parmak izlerimiz birbirinden farklıdır.

bireysel farklılıklara saygı duyulması insanlar birbirlerini asalanaya başlar ülkede iç savaş çıkabilir mesela

3 0
3 years ago
Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp
exis [7]

Answer:

H_0: \, \mu = 2.

H_1:\, \mu > 2.

Test statistics: z \approx 2.582.

Critical value: z_{1 - 0.025} \approx 1.960.

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean \mu is greater than 2. This claim should be reflected in the alternative hypothesis:

H_1:\, \mu > 2.

The corresponding null hypothesis would be:

H_0:\, \mu = 2.

In this setup, the null hypothesis H_0:\, \mu = 2 suggests that \mu_0 = 2 should be the true population mean of GPA.

However, the alternative hypothesis H_1:\, \mu > 2 does not agree; this hypothesis suggests that the real population mean should be greater than \mu_0= 2.

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed \mu_0 = 2 (i.e., the null hypothesis is true.) How likely would the sample (sample mean \overline{X} = 2.02 with sample standard deviation s = 0.06) be observed in this hypothetical population?

Let \sigma denote the population standard deviation.

Given the large sample size n = 60, the population standard deviation should be approximately equal to that of the sample:

\sigma \approx s = 0.06.

Also because of the large sample size, the central limit theorem implies that Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} should be close to a standard normal random variable. Use a Z-test.

Given the observation of \overline{X} = 2.02 with sample standard deviation s = 0.06:

\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}.

Because the alternative hypothesis suggests that the population mean is greater than \mu_0 = 2, the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed z-test. The question requested a significant level of 0.025. Therefore, the critical value z_{1 - 0.025} should ensure that P( Z > z_{1 - 0.025}) = 0.025.

Look up an inverse Z table. The z_{1 - 0.025} that meets this requirement is z_{1 - 0.025} \approx 1.960.

The z-value observed from the sample is z_\text{observed}\approx 2.582, which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this 0.025 confidence level in favor of the alternative hypothesis.

3 0
3 years ago
Linda drives 35 miles five days of the week . During one year, she has 10 days of vacation and 10 holidays when she does not dri
Alenkinab [10]

Answer:

1680 miles

Step-by-step Explanation:

There are 52 weeks in a year. Total number of days she is expected to work would be 52 * 5 = 260 days.

Now we know she has 10 vacation days and 10 holidays. This translates 260 - 20 = 240 days

Now if she drives 35miles 5 days of the week, this means she dives 7 miles per day.

The total number of miles traveled is this 240 * 7 = 1680 miles

5 0
3 years ago
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Please help me please​
creativ13 [48]

Answer:

i dont know

Step-by-step explanation:

3 0
3 years ago
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