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Yuliya22 [10]
3 years ago
6

Mr. D'Agostino decided he wanted to build a pool in his backyard this summer. He sketched a drawing that had a scale of 1 in. =

4 feet. If the drawing had a length of 4 feet and a width of 5 feet, what is the actual area of the pool?
Mathematics
1 answer:
Travka [436]3 years ago
5 0

Answer:

46080 ft²

Step-by-step explanation:

Given :

Scale :

1 in = 4 feets

Recall :

I foot =. 12 inches

Length of drawing = 4 feets

Length of drawing in inches = 4 * 12 = 48 inches

Width of drawing = 5 feets

Width of drawing in inches = 5 * 12 = 60 inches

Actual length of pool = 48 * 4 = 192 feets

Actual width of pool = 60 * 4 = 240 feets

The actual area of the pool :

Length * width

192 feets * 240 feets = 46080 ft²

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Find the greatest common factor and the least common multiple of 16 and 20. The prime factorizations of each number are given.
kykrilka [37]

Answer:

GCF 4, LCM 80

Step-by-step explanation:

Since both 16 and 20 have two 2s as factor, their greatest common factor is 4.

The LCM is found by multiplying all of the remaining factors by the LCM:

16 still has (2 x 2), 20 still has (5), times the GCF (4)  so 2 x 2 x 5 x 4 = 80.

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2 years ago
Mr. Duncan bought a table at a discount if 30% thus saving $42. what was the marked price of the table?​
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Answer:

$140

Step-by-step explanation:

if 30% of full price = $42 then 100% = 42/0.3 = 140

5 0
3 years ago
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Factor the following expression using the GCF: 10ab + 20.
Lostsunrise [7]
The first one is 10ab + 20
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3 years ago
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The charge is $12 plus $0.15 per tree. What is the greatest number of trees that can be planted if you spend no more than $70
notsponge [240]

Answer:

386

Step-by-step explanation:

If you subtract the initial fee of 12 from 70 you get 68 you just divide that by .15 meaning you can plant no more that 386 trees.

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A recent broadcast of a television show had a 10 ​share, meaning that among 6000 monitored households with TV sets in​ use, 10​%
Kisachek [45]

Answer:

Null and alternative hypothesis

H_0: \pi \geq0.25\\\\H_1: \pi

Test statistic z=-26.82

P-value P=0

The null hypothesis is rejected.

It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

Step-by-step explanation:

We have to perform a hypothesis test of a proportion.

The claim is that less than 25% were tuned into the program, so we will state this null and alternative hypothesis:

H_0: \pi \geq0.25\\\\H_1: \pi

The signifiance level is 0.01.

The sample has a proportion p=0.1 and sample size of n=6000.

The standard deviation of the proportion, needed to calculate the test statistic, is:

\sigma_p=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.25(1-25)}{6000} } =0.0056

The test statistic is calculated as:

z=\frac{p-\pi+0.5/N}{\sigma_p} =\frac{0.1-0.25+0.5/6000}{0.0056}=\frac{-0.1499}{0.0056}  =-26.82

As this is a one-tailed test, the P-value is P(z<-26.82)=0. The P-value is smaller than the significance level (0.01), so the effect is significant.

Since the effect is significant, the null hypothesis is rejected. It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

3 0
3 years ago
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