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3 years ago

What is the area of a circle that has circumference of 42 inches

Mathematics

1 answer:

mixer [17]3 years ago

7 0

I think the area of a circle with a circumference of forty two inches is
140.37 in^2

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In psychology, there is a particular Mental Development Index (MDI) used in the study of infants. The scores on the MDI have app

ivann1987 [24]

Answer:

0.1587

Step-by-step explanation:

Here, mean=μ=100 and standard deviation=σ=16.

We have to find P(average MDI scores of 64 children > 102)=P(xbar>102).

n=64

μxbar=μ=100

σxbar=σ/√n=16/√64=16/8=2

P(xbar>102)=P((xbar-μxbar)/σxbar>(102-100)/2)

P(xbar>102)=P(z>1)

P(xbar>102)=P(0<z<∞)-P(0<z<1)

P(xbar>102)=0.5-0.3413

P(xbar>102)=0.1587

Thus, the probability that the average is greater than 102 is 15.87%

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3 years ago

I WILL MARK U AS A BRAINLIEST! Please help question on the picture.

tester [92]

Answer:

just wanted to let you know you didn't attach a picture would love to answer question in comments once I can see the photo

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3 years ago

Mrs. Varner deposited q dollars

Marizza181 [45]

Answer:

this is not a full question

Step-by-step explanation:

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2B%20%20%5Cfrac%7Bw%7D%7B3%7D%20" id="TexFormula1" ti

ElenaW [278]

Answer: $\bold{w=\dfrac{6y-3}{2}}$

<u>Step-by-step explanation:</u>

Isolate w by performing the following steps

Multiply by 6 on both sides to clear the denominator
Subtract 3 from both sides
Divide both sides by 2

$y=\dfrac{1}{2}+\dfrac{w}{3}\\\\\\6\bigg[y=\dfrac{1}{2}+\dfrac{w}{3}\bigg]\quad \implies \quad 6y=3+2w\\\\\\6y-3=3-3+2w\quad \implies \quad 6y-3=2w\\\\\\\dfrac{6y-3}{2}=\dfrac{2w}{2}\quad \implies \quad \large\boxed{\dfrac{6y-3}{2}=w}$

5 0

3 years ago

100 POINTS AND BRAINLIEST

zysi [14]

Answer:

8 square units and $\frac{40}{3}$ square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor $\frac{1}{3},$ then

$\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.$

2. Triangles ABC and DBE are similar with scale factor $\frac{2}{3},$ then

$\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.$

3. Thus, the area of the quadrilateral DFGE is

$A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.$

and the area of the quadrilateral ADEC is

$A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.$

4 0

3 years ago

Read 2 more answers