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Talja [164]
3 years ago
15

The quotient of a number and 14 is 8. Find the number.​

Mathematics
2 answers:
frosja888 [35]3 years ago
6 0

Answer:

112

Step-by-step explanation:

Take the missing number as x.

Then the required value of x will be,

→ x ÷ 14 = 8

→ x = 8 × 14

→ [ x = 112 ]

Hence, the missing number is 112.

mafiozo [28]3 years ago
5 0

Answer:

The number is 112

Step-by-step explanation:

Let the number be 'x'

\sf \dfrac{x}{14}=8\\\\\text{Multiply both sides by 14}\\\\ \dfrac{x}{14}*14=8*14\\\\

<h3>          x  = 112</h3>
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What is 6 times 10 to the second negative power and 4.2 times to the the negative 3rd power
OverLord2011 [107]
Answer:

6 * 10 to the -2nd power = 0.00027777777

4.2 to the -3rd power = 0.01349746247

If this helped leave a thanks, thanks.
4 0
3 years ago
A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute
GREYUIT [131]

Answer:

(a) Null Hypothesis, H_0 : \sigma = 10 beats per minute  

     Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute  

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal​ range, the result is a standard deviation of 10 beats per minute.

Let \sigma = <u><em>population standard deviation for the pulse rates of men</em></u>.

(a) So, Null Hypothesis, H_0 : \sigma = 10 beats per minute      {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute      {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is <u>One-sample chi-square test</u> for standard deviation;

                             T.S.  =  \frac{(n-1)\times s^{2} }{\sigma^{2} }  ~  \chi^{2}__n_-_1  

where, s = sample standard deviation = 10.1 beats per minute

            n = sample of men = 36

So, <u><em>the test statistics</em></u> =  \frac{(36-1)\times 10.1^{2} }{10^{2} }  ~ \chi^{2}__3_5

                                   =  35.70  4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

                    P-value = P(\chi^{2}__3_5 > 35.704) = <u>0.4360</u>

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

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