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Ksju [112]
3 years ago
10

Trent is fishing from a pier. ●The tip of his fishing rod is 53 ¾ feet above the surface of the water. ●The hook on the end of t

he fishing line is directly below the tip of the fishing rod 12 ⅔ feet below the surface of the water. Trent estimates that the distance between the tip of his fishing rod and the hook is less than 65 feet. Is Trent’s estimate reasonable? Explain your answer. HELP Please
Mathematics
1 answer:
iris [78.8K]3 years ago
6 0
I would say yes and no because estimations aren't exact, they're close to the correct value. Trent's estimation is correct since 53+12=65, if we were looking for the exact distance, Trent would be wrong because there are fractions too, so the exact value would be 66 (5/12) or 65 (17/12).

Hope this helped :)
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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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2 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
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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
Find the slope of the line. y = 5/3x + 4
ioda

Answer:

slope= 5/3

Step-by-step explanation:

y=mx+b

slope=m

your welcome

7 0
3 years ago
Read 2 more answers
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