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tatuchka [14]
3 years ago
5

What is the solution

Mathematics
1 answer:
Tju [1.3M]3 years ago
7 0

Answer:

m =40

Step-by-step explanation:

m/4 -3 = 7

Add 3 to each side

m/4 -3+3 = 7+3

m/4 = 10

Multiply each side by 4

m/4 *4 = 10*4

m = 40

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The circumference is about ___ m <br><br>use 3.14 for pi.​
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The circumference is:

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the sum of total amount of money is distributed equally to some girl If there has been five girls less than each of them get Rs.
Arturiano [62]

Step-by-step explanation:

Let the sum of rupee be x

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3 years ago
Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o
kirza4 [7]

Answer:

a) H0: \sigma = 1.34

H1: \sigma \neq 1.34

b) df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

c) t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

\sigma_o =1.34 the value that we want to test

p_v represent the p value for the test

t represent the statistic  (chi square test)

\alpha=0.01 significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: \sigma = 1.34

H1: \sigma \neq 1.34

The statistic is given by:

t=(n-1) [\frac{s}{\sigma_o}]^2

Part b

The degrees of freedom are given by:

df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

Part c

Replacing the info we got:

t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0
3 years ago
prove that if f is a continuous and positive function on [0,1], there exists δ &gt; 0 such that f(x) &gt; δ for any x E [0,1] g
ValentinkaMS [17]

Answer:

I dont Know

Step-by-step explanation:

5 0
3 years ago
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