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Oduvanchick [21]
3 years ago
7

WILL MARK BRAINLIEST + 10 POINTS #14

Mathematics
1 answer:
3241004551 [841]3 years ago
5 0
This question is asking "Where does cosine equal \frac{\sqrt{3}}{2}?"

Based on our unit circle values and the domain of the cos^{-1} function, we see that at 30 degrees, or \frac{\pi}{6}, cosine is equal to \frac{\sqrt{3}}{2}.
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Pls help me........ I tried lol
diamong [38]

its step 2 he did not subtract it right hope this helps and have a great day and get some sleep lol

7 0
2 years ago
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PLEASE HELP THANK YOU!
Vsevolod [243]

Answer:

um wow that is realy hard

Step-by-step explanation:

8 0
3 years ago
Jennifer began an exercise program in which she ran 4 blocks the first day and then two blocks more than the previous day for ea
Andrews [41]

Answer:

84 blocks.

Step-by-step explanation:

If you are adding 2 to each day (which already has 4) in a 2 week period (14 days) than you would have to time 2(14) and 4(14) and add them. Thus would make the answer 84 blocks.

5 0
2 years ago
One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla
Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0
3 years ago
6+5(9÷3)2 help answer it pls
lorasvet [3.4K]

Answer:

The answer should be 36, unless there is a typo

8 0
3 years ago
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