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ycow [4]
3 years ago
11

Square A is a scaled version of square B. The dimensions of square A are three times the dimensions of square B. The area of squ

are A is 900 sq cm.
Mathematics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

The side of square A is 30 cm.

Step-by-step explanation:

Given that,

The dimensions of square A are three times the dimensions of square B.

The area of square A is 900 sq cm.

Side of square A = 3( side of square B)

Let the side of square B is b. So,

Side of square A, a = 3b

We know that, the area of square is side². So,

(3b)^2=900\\\\9b^2=900\\\\b^2=100\\\\b=10\ cm

Side of square A, a = 3(10) = 30 cm

Hence, the side of square A is 30 cm.

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the answer is to 62-6.75 is 55.25
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a half cup of pancakes mix has 5% total daily allowance of cholesterol. the total daily allowance of cholesterol 300 mg. how muc
arsen [322]
Hi there!

Since we know that the pancake mix has 5% of the daily allowance, which is 300 mg, we can just multiply 300 mg with 5 percent to get our answer.

First, let's divide 5% by 100% to know what we are going to multiply 300 mg with.

5% / 100% = 0.05

Let's multiply 300 mg with 0.05

0.05 * 300 = 15 mg

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3 years ago
4log(2) x + log(2)5 = log(2)405
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Answer:

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Step-by-step explanation:

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Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t
Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let \sigma_{1}^{2} be the variance for the population of weight gains for rats given a low dose, and \sigma_{2}^{2} the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2} vs H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}. We have that the sample standard deviation for n_{2} = 22 female control rats was s_{2} = 28 g and for n_{1} = 18 female low-dose rats was s_{1} = 51 g. So, we have observed the value

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As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

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3 years ago
Simplify (5b10)(−2b−3).
Angelina_Jolie [31]

Answer:

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Step-by-step explanation:

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