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Sergio [31]
4 years ago
7

A pipe that's 36 inches long needs to be cut into 2 and one forth inches long pieces. how may pieces can be cut from this length

of pipe
Mathematics
1 answer:
PolarNik [594]4 years ago
7 0
16 pieces may be cut. This is because we need to divide 36 by 2 and 1/4 to find how many pieces hat can be cut. First, we need to change everything into a fraction (or improper fraction). 2 and 1/4 = 9/4 (if you need help trying to find out how to get this, just ask). Then, since we are doing division, we flip the 9/4 around to get 4/9. 36 times (it's not dividing anymore since we flipped the 9/4) 4/9 = 16 pieces. I hope this helps! :D
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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

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Answer:

On EDG the answer is D

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

5*4=70

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When the positive integer n is divided by 7, the remainder is 2. What is the remainder when 5n is divided by 7?
ser-zykov [4K]

Therefore the reminder is 3 when 5n is divided by 7.

Step-by-step explanation:

we know that

Dividend = quotient × Divisor +Reminder

⇔ n = Q(n) × 7 + 2          [ let Q(n)  is quotient]

multiplying by 5 both sides

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⇔5n = 5 Q(n) ×7 +10

⇔5n=  5 Q(n) ×7 +7 +3

⇔ 5n = (5Q(n) +1) × 7 +3

Therefore the reminder is 3 when 5n is divided by 7.

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Answer:

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Step-by-step explanation:

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