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vodomira [7]
2 years ago
6

31. The angle measures of a triangle are 28°, 70°, and 82°. Classify the triangle by its angle measures.

Mathematics
1 answer:
ZanzabumX [31]2 years ago
4 0

Answer:

Acute scalene triangle

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sergij07 [2.7K]

For this case we have the following arithmetic sequence:

a_ {n} = 15 + 5 (n-1)

To write in function form, we apply distributive property to the terms within parentheses:

f (n) = 15 + 5n-5

Different signs are subtracted and the major sign is placed.

We simplify:

f (n) = 5n + 10

Answer:

f (n) = 5n + 10

Option C

4 0
3 years ago
Find the value of x plz help
Rzqust [24]

Answer:

29 degrees

Step-by-step explanation:

Both angles are the same.

8 0
3 years ago
Read 2 more answers
Historically, a certain region has experienced 92 thunder days annually. (A "thunder day" is day on which at least one instance
DiKsa [7]

Answer:

We conclude that the mean number of thunder days is less than 92.

Step-by-step explanation:

We are given that Historically, a certain region has experienced 92 thunder days annually.

Over the past fifteen years, the mean number of thunder days is 72 with a standard deviation of 38.

<u><em>Let </em></u>\mu<u><em> = mean number of thunder days.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 92 days     {means that the mean number of thunder days is more than or equal to 92}

Alternate Hypothesis, H_A : \mu < 92 days     {means that the mean number of thunder days is less than 92}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean number of thunder days = 72

             s = sample standard deviation = 38

            n = sample of years = 15

So, <u><em>test statistics</em></u>  =  \frac{72-92}{\frac{38}{\sqrt{15} } }  ~ t_1_4

                               =  -2.038

The value of z test statistics is -2.038.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.

Since our test statistics is less than the critical value of t as -2.038 < -1.761, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean number of thunder days is less than 92.

7 0
3 years ago
Which of the following equations represents an inverse variation? A. 6y = x B.
faltersainse [42]
The answer is 6x xb is correct
3 0
3 years ago
Pouvez vous maider pour la questuon deux merci
Vanyuwa [196]

quelle est ta question?


6 0
3 years ago
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