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1 year ago

I have 90% after about 75%

Mathematics

1 answer:

kkurt [141]1 year ago

6 0

The lowest grade he could earn is 120% of the grade at the end of the semester.

The following statement is given:

I have 90% after about 75% of the semester.

We are asked to find the lowest grade by the end of this semester.

<h3>What is Percentage?</h3>

A percentage is a number expressed as a fraction of 100.

- 50% = 50/100 = 1/2

- 25% = 25/100 = 1/4

- 20% = 20/100 = 1/5.

We can write this statement "I have a 90% grade after about 75% of the semester" as:

90% grade = 75% semester.............(1)

By the end of the semester means at 100% semester.

Multiplying equation (1) by 100/ 75 on both sides of the equation.

We get,

(100/75) x 90% grade = (100/75) x 75% semester

(100 x 90)/75 % grade = 100% semester

120% grade = 100%

Thus the lowest grade he could earn is 120% of the grade at the end of the semester.

Learn more about Percentages here:

brainly.com/question/24159063

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<h3>What is a confidence interval of proportions?</h3>

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a) $z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

b) $p_v =2*P(z$

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Step-by-step explanation:

Information given

$\bar X_{1}= 104$ represent the mean for 1

$\bar X_{2}= 106$ represent the mean for 2

$\sigma_{1}= 8.4$ represent the population standard deviation for 1

$\sigma_{2}= 7.6$ represent the population standard deviation for 2

$n_{1}=80$ sample size for the group 1

$n_{2}=70$ sample size for the group 2

z would represent the statistic

Hypothesis to test

We want to check if the two means for this case are equal or not, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

The statistic would be given by:

$z =\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma^2_1^2}{n_1} +\frac{\sigma^2_2^2}{n_2}}}=$ (1)

Part a

Replacing we got:

$z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

Part b

The p value would be given by this probability:

$p_v =2*P(z$

Part c

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