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GrogVix [38]
3 years ago
15

Change 4 and 3/5 to an improper fraction​

Mathematics
1 answer:
atroni [7]3 years ago
7 0

Answer:

The answer is 23/5, or 23 over 5.

Step-by-step explanation:

4 and 3/5

Multiply the denominator by the whole number and add the numerator, all over the original denominator to calculate the improper fraction.

20 + 3 over 5

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2. What is the slope of the line through the points (-6, 3) and (-7,-4)?
Gekata [30.6K]

Answer:

The answer is d man .....

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4 years ago
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kurt spots a bird sitting at the top of a 40 foot tall telephone pole. If the angle of elevation from the ground where he is sta
Likurg_2 [28]

angle of elevation, aoe = 59 deg

height, h = 40ft

distance from pole, dfp

using

tan 59deg = h/dfp

=》 dfp = h/ tan 59deg = 40/1.67 = 23.95ft

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3 years ago
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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
2/7+2/5=answernin simplest form
MrRa [10]

Answer:24/34

Step-by-step explanation:

5 0
4 years ago
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1. 10:15 H Wednesday (Philippines) - 05:15 H Wednesday (Riyadh, Saudi<br>Arabia)<br>Time difference​
kotegsom [21]

Answer:

5hours

Step-by-step explanation:

10:15 - 05:15 = 5 hours

Thsi means that local time in Philippines is 5 hours ahead of local time in Riyadh, Saudi Arabia.

The time can be interpreted as follows ; At a certain time in the Philippines, say 10:15 am in the morning, At this verh point in the Riyadh, Saudi Arabia, the time will be :

(10:15 - 5 hours) = 05:15

7 0
3 years ago
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