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3 years ago

Triangle ABC has area 45cm. Find the length AB. 10cm 400

Mathematics

1 answer:

nekit [7.7K]3 years ago

5 0

Answer:

if the triangle is equilateral, then side AB would measure 15cm

Step-by-step explanation:

side AB would be 15cm, because the area is 45 cm and there are 3 sides, so 45/3=15. again, assuming the triangle is equilateral

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All Major League Baseball stadiums use standard measurements for their infield area. The pitcher's mound is a circle with a diam

denis-greek [22]

Answer:

254 4/7 or 254.34 ft² (it depends on which method you use.)

Step-by-step explanation:

To find the area of a circle, the formula is πr².

I'll use both 22/7 and 3.14 as pi, so I'll end up with two different answers. Just choose the more reliable one.

22/7 version:

22/7 * 9^2

22/7 * 81/1

1782/7

254 4/7 ft²

3.14 version:

3.14 * 9^2

3.14 * 81

314 - (3.14 * 19)

314 - 62.8 + 3.14

251.2 + 3.14

254.34 ft²

As I said, I ended up with two different answers. You also had said not to round my answer, so the 22/7 version has a mixed number.

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3 years ago

A parallelogram has an area of 60 square inches. If the base of the parallelogram is 12 inches, what is the height of the parall

Vikentia [17]

How to find the area of a parallelogram is length times height. 12 times what gets you to 60? 5 does. 12 times 5 is 60.

HEIGHT: 5

8 0

3 years ago

Select two ratios that are equivalent to 4:18

fredd [130]

36:8, 9:2, and 72: 16

Step-by-step explanation:

two ratios that are equivalent to 4:18

36:8, 9:2, and 72: 16

would all be equivalent. In order to find ratios that are equivalent to 18:4, you have to multiply both numbers by the same factor. For

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3 years ago

What is the y-intercept of line MN?

Vlad1618 [11]

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3 years ago

Read 2 more answers

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

uysha [10]

Answer:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

Step-by-step explanation:

For this problem we have the following data given:

$\bar X_1 = 9$ represent the sample mean for one of the departments

$\bar X_2 = 8$ represent the sample mean for the other department

$n_1 = 25$ represent the sample size for the first group

$n_2 = 20$ represent the sample size for the second group

$s_1 = 2$ represent the deviation for the first group

$s_2 =1$ represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

The confidence given is 95% or 9.5, then the significance level is $\alpha=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n_1 +n_2 -2= 20+25-2= 43$

And the critical value for this case is $t_{\alpha/2}=2.02$

And replacing we got:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

4 0

2 years ago