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Elza [17]
3 years ago
15

What is 576 divided by 47

Mathematics
1 answer:
viktelen [127]3 years ago
4 0

The answer to your question,

If rounded it would be 12.3, if not 12.5 with a repeated.

-Mabel

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Dima020 [189]

Answer:

Question A is D and Question B is B

3 0
3 years ago
Read 2 more answers
What is the next term in the sequence 1/2 1/6 1/18 1/54
ddd [48]
1/162
2*3=6
6*3=18
18*3=54
54*3=162
7 0
2 years ago
When 2 is added to the difference between six times a number and 5, the result is greater than 13 added to 5 times the number. F
Elina [12.6K]

let x is the number

2 + (6x - 5) > 13 + 5x

2 + 6x - 5 > 13 + 5x

6x - 3  > 13 + 5x

6x - 5x > 13 + 3

         x >  16

8 0
3 years ago
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
Solve the Equation (NO LINKS AND ONLY IF YOU KNOW PLS)
Dafna1 [17]
O.573994929277273994
4 0
3 years ago
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