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3 years ago

5

the area of triangle b is 64 times greater than the area of triangle a.the perimeter of triangle b is how many times greater tha

n the perimeter of triangle a?

1 answer:

aev [14]3 years ago

8 0

So yah.. perimeter of triangle b is
= 32 + a/2

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hoa [83]

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Step-by-step explanation:

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Samantha measured the dimensions of a rectangular poster frame. It was 3/4 yard long and 1/2 yard wide. What is the area of the

Damm [24]

$\frac{3}{8} {yards}^{2}$
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4 years ago

Read 2 more answers

The average of Johnny's two test grades is a 92 what must he get on his 3rd test to raise his average to a 94?

jeka57 [31]

Answer:

Johnny needs to score 98 on third test to raise average to 94

Step-by-step explanation:

Given that average of two tests is 92.

The score of next test has to be found that means there will be a total of 3 tests.

The average will be:

$Avg = \frac{test1+test2+test3}{3}$

It is also given that he has to raise his average to 98

We have to use the average formula for finding the third value

So,

$94 = \frac{92+92+test3}{3}\\94*3 = 184+test3\\282 = 184+test3\\test3 = 282-184\\test3 = 98$

Hence,

Johnny needs to score 98 on third test to raise average to 94

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3 years ago

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The combined (verbal + quantitative reasoning) score on the GRE (Graduate Record Exam) is normally distributed with mean 1049 an

mash [69]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Normal distribution :

Mean (m) = 1049

Standard deviation (sd) = 189

Number of samples (n) = 16

A) What is the probability that one randomly selected student scores above 1100 on the GRE?

Obtain the z score

Zscore = (x - m) / sd

Zscore = (1100 - 1049) / 189

Zscore = 51 / 189 = 0.2698412 = 0.27

Using the z-table

P(X > 100) = 1 - P(X < 100) = 1 - 0.6064 = 0.3936

B) The sample mean = population mean

Hence sample mean = 1049

The sample standard deviation = standard error

Standard Error (SE) = sd / sqrt(n)

SE = 189/4 = 47.25

Hence shape = normal, standard deviation = 47.25, mean = 1049

C.) P(X < 1100)

Zscore = (x - m) / SE

Zscore = (1100 - 1049) / 47.25

Zscore = 51 / 47.25 = 1.0793650 = 1.08

Using the z-table :

P(X < 1100) = 0.8599

D.) P(X > 1100)

Zscore = (x - m) / SE

Zscore = (1100 - 1049) / 47.25

Zscore = 51 / 47.25 = 1.0793650 = 1.08

Using the z-table :

P(X > 1100) = 1 - P(X < 1100) = (1 - 0.8599) = 0.1401

E) if n = 64

SE = 189 / sqrt(64) = 189 / 8 = 23.625

P(X > 1100)

Zscore = (x - m) / SE

Zscore = (1100 - 1049) / 23.625

Zscore = 51 / 47.25 = 2.1587301 = 2.16

Using the z-table :

P(X > 1100) = 1 - P(X < 1100) = (1 - 0.9846) = 0.0154

The probability decreases as the same distribution hinges towards a Boal distribution with increasing sample size.

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3 years ago