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Lesechka [4]
2 years ago
11

Trevon, Leah, and Beth are playing a computer game. Trevon scores -8. Beth’s score was 3/4 of trevons score, and Leah’s score wa

s 1/4 of Beth’s score. What was Leah’s score
Mathematics
2 answers:
FrozenT [24]2 years ago
8 0

Answer:

-1.5

Step-by-step explanation:

We have been given that Beth's score is 3/4 of Trevon's score.

\text{Beth's score}=-8\times \frac{3}{4}

\text{Beth's score}=-2*3

\text{Beth's score}=-6

We are also told that Leah’s score was 1/4 of Beth’s score.

\text{Leah's score}=-6\times \frac{1}{4}

\text{Leah's score}=-\frac{3}{2}

\text{Leah's score}=-1.5

Therefore, Leah score was -1.5.

nevsk [136]2 years ago
3 0
-8 = trevon

(3/4)-8 = beth
= -6

1/4[(3/4)-8] = leah
1/4(-6)
= -1.5
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After 22 seconds the projectile reach its maximum height of 4,840 units

Step-by-step explanation:

we have

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Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and
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Answer:

t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349      

p_v =P(t_9>10.349)=1.34x10^{-6}  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation      

\bar X=51.6 represent the sample mean

s=1.1 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =48 represent the value that we want to test    

\alpha=0.05 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.      

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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{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349      

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