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SpyIntel [72]
4 years ago
11

Pls help I need this done by today!

Mathematics
2 answers:
pochemuha4 years ago
4 0

Answer:

u = -2

Step-by-step explanation:

-6 + 3u = -12

Add six to both sides

3u = -6

Divide out the 3

u = -2

slava [35]4 years ago
3 0

Answer:

U= -2

Step-by-step explanation:

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Is -15.51+ 8.55 positive or negative​
kherson [118]

Answer:

negative

Step-by-step explanation:

4 0
3 years ago
Solve the equation: 5 ( x - 2) + 4 ( 3 + x) = 20​
bixtya [17]

Answer:

x=2

Step-by-step explanation:

5 ( x - 2) + 4 ( 3 + x) = 20​

Distribute

5x -10 +12 +4x = 20

Combine like terms

9x +2 = 20

Subtract 2 from each side

9x+2-2 = 20-2

9x = 18

Divide each side by 9

9x/9 = 18/9

x = 2

3 0
3 years ago
Read 2 more answers
6 points and brainliest answer
kati45 [8]
Well, the angle looks like it is an obtuse angle, right? And we know that obtuse angles are greater than 90 degrees, so wouldn't the answer be A. 115 degree?
I think the answer is 115 degrees. Hope this helps!
4 0
3 years ago
Read 2 more answers
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
2 over 7x +8=22 solve for x
yanalaym [24]
In this question the answer is x is equal to exactly 1.749... Or 1.75

4 0
3 years ago
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