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3 years ago

5 5/8 - 5 1/8 simplify .

Mathematics

1 answer:

uranmaximum [27]3 years ago

4 0

Answer: 5 : 3 1/8 = 8 : 5

Step-by-step explanation:

Change values to whole numbers.

Convert any mixed numbers to fractions.

Convert 3 1/8

3 1/8 = 25/8

We now have:

5 : 3 1/8 = 5 : 25/8

Convert the whole number 5 to a fraction with 1 in the denominator.

We then have:

5 : 3 1/8 = 5/1 : 25/8

Convert fractions to integers by eliminating the denominators.

Our two fractions have unlike denominators so we find the Least Common Denominator and rewrite our fractions as necessary with the common denominator

LCD(5/1, 25/8) = 8

We now have:

5 : 3 1/8 = 40/8 : 25/8

Our two fractions now have like denominators so we can multiply both by 8 to eliminate the denominators.

We then have:

5 : 3 1/8 = 40 : 25

Try to reduce the ratio further with the greatest common factor (GCF).

The GCF of 40 and 25 is 5

Divide both terms by the GCF, 5:

40 ÷ 5 = 8

25 ÷ 5 = 5

The ratio 40 : 25 can be reduced to lowest terms by dividing both terms by the GCF = 5 :

40 : 25 = 8 : 5

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Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

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Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

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