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Annette [7]
3 years ago
8

Marta drew a scale drawing of her backyard. She used the scale 1/2 inch =4.5 feetif the actual length of her backyard is 54,what

is her scale drawing in inches
Mathematics
1 answer:
RoseWind [281]3 years ago
8 0

1/2 inch…………4.5 feet

x inches……….54 feet

x=(54•1/2)/4.5

x=27/4.5

x=6 inches


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Answer:

A

Step-by-step explanation:

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3 years ago
In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were re
11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μemployed(μ1) be the sample mean of students who are employed and μnot employed(μ2) be the sample mean of students who are not employed

The random variable is μ1 - μ2 = difference in the mean of the employed and unemployed students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 3.22

μ2 = 3.33

s1 = 0.475

s2 = 0.524

n1 = 172

n2 = 116

t = (3.22 - 3.33)/√(0.475²/172 + 0.524²/116)

t = - 1.81

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.475²/172 + 0.524²/116]²/[(1/172 - 1)(0.475²/172)² + (1/116 - 1)(0.524²/116)²] = 0.00001353363/0.00000005878

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We would determine the probability value from the t test calculator. It becomes

p value = 0.036

Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis.

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3 0
3 years ago
If three pears and four oranges cost $4.85 and three pears and ten oranges cost $8.75 what is the cost of a pear and an orange
marishachu [46]

Answer:

1 pear = $0.75; 1 orange = $0.65

Step-by-step explanation:

(1)               3P +  4O = 4.85

(2)              3P + 10O = 8.75                 Eqn (2) - Eqn (1)

3P + 10O – 3P – 4O = 8.75 – 4.85     Combine like terms

                           6O = 3.90                 Divide each side by 6

                             O = $0.65              Substitute into Eqn (1)

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               3P + 2.60 = 4.85                 Subtract 2.60 from each side

                           3P = 2.25                Divide each side by 3

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1 is in the tens place

we look at 2

2 <5 so we leave it alone

6,310

Choice B

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