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3 years ago

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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

2 answers:

FrozenT [24]3 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]3 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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Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

a

$P(\= X \ge 51 ) =0.0062$

b

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

So

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

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