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4 years ago

If the area of BAC=8, what is the area of EDF. Enter a decimal rounded to the tenths.

Mathematics

1 answer:

777dan777 [17]4 years ago

7 0

Answer:

4.5 in²

Step-by-step explanation:

Given that the triangles are similar then

linear ratio of corresponding sides = 3 : 4, then

ratio of corresponding areas = 3² : 4² = 9 : 16

let the area of ΔEDF be x then by proportion

$\frac{x}{9}$ = $\frac{8}{16}$ ( cross- multiply )

16x = 72 ( divide both sides by 16 )

x = 72 ÷ 16 = 4.5

Area of ΔEDF = 4.5 in²

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Write an equation to show the relationship between a and c when a = b + 9 b = c - 2

ICE Princess25 [194]

Given:

The equations are:

$a=b+9$

$b=c-2$

To find:

The equation that show the relationship between a and c.

Solution:

We have,

$a=b+9$ ...(i)

$b=c-2$ ...(ii)

Equation (i) can be rewritten as:

$a-9=b$ ...(iii)

From (ii) and (iii), we get

$a-9=c-2$

$a-c=9-2$

$a-c=7$

Therefore, the required equation is $a-c=7$ .

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3 years ago

crystal is picking blueberries.so far,she has filled 2/3 of her basket,and the blueberries weigh 3/4pound .the equation 2/3w=3/4

vesna_86 [32]

In order to do this, we multiply both sides be the amount that would make the left side equal to 1. This number is the reciprocal of the fraction, so we multiply both sides by 3/2. This is equal to:

w = 9/8

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3 years ago

This is question is =5* 0.25

Kruka [31]

I am new here

Answer: 1.25

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3 years ago

Read 2 more answers

Which are the ways to finance your college education?

scZoUnD [109]

Answer:

All of the above

Step-by-step explanation:

Grants. Colleges, states, and the federal government give out grants, which don't need to be repaid.

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3 years ago

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Assume that both populations are normally distributed. a) Test whether mu 1 not equals mu 2 at the alpha equals 0.05 level of s

Radda [10]

Answer:

$(16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445$

$(16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$n_1 = 19$ sample size 1

$n_2 = 19$ sample size 2

$\\bar X_1 = 16.2$ sample mean for group 1

$\\bar X_2 = 14.2$ sample mean for group 2

$s_1 = 4.5$ sample deviation for group 1

$s_2 = 3.1$ sample deviation for group 2

Solution to the problem

For this case the confidence interval is given by:

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

And the degrees of freedom are given by:

$df = n_1 +n_2 -2 = 19+19 -2= 36$

We want a 95% of confidence o then the significance level is 1-0.95 =0.05 and $\alpha/2 = 0.025$ if we find a critical value in the t distribution with 36 degrees of freedom we got:

$t_{cric} =2.03$

And replacing we got:

$(16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445$

$(16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645$

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3 years ago