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3 years ago

A teacher has 36 students who were randomly assigned to 6 rows of 6 desks. She wants to survey the students

Mathematics

1 answer:

ivanzaharov [21]3 years ago

5 0

Answer:

Step-by-step explanation:

Edge 2021

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The length of a rectangle is twice its width. Find its area, if its perimeter is 7 1/3 cm The area of the rectangle is _sq. cm.

Harman [31]

Answer:

The area of rectangle is $2\frac{80}{81}\ cm^2$

Step-by-step explanation:

we know that

The perimeter of a rectangle is equal to

$P=2(L+W)$

we have

$P=7\frac{1}{3}\ cm=7+\frac{1}{3}=\frac{22}{3}\ cm$

$\frac{22}{3}=2(L+W)$

$\frac{11}{3}=(L+W)$ ----> equation A

Remember that

The length of a rectangle is twice its width.

$L=2W$ ----> equation B

substitute equation B in equation A

$\frac{11}{3}=(2W+W)$

solve for W

$W=\frac{11}{9}\ cm$

Find the value of L

$L=2(\frac{11}{9})$

$L=\frac{22}{9}\ cm$

<em>Find the area of rectangle</em>

$A=LW$

substitute the values

$A=(\frac{22}{9})(\frac{11}{9})$

$A=\frac{242}{81}\ cm^2$

Convert to mixed number

$\frac{242}{81}\ cm^2=\frac{162}{81}+\frac{80}{81}=2\frac{80}{81}\ cm^2$

7 0

3 years ago

What is the area of the figure ?

Sliva [168]

The area is 48

I did 12x8/2 using the formula height x base divided by 2

hope it helps

7 0

3 years ago

Suppose a NBA basketball player has 5% chance of picking up a foul every minute he plays. What is the probability that he will p

alukav5142 [94]

Answer:

He wont foul out with probability 0.9093

Step-by-step explanation:

The total number of fools he picked is a Binomial ditribution noted by X with parameters p = 0.05 and N = 48. The mean of this random variable is μ = np = 48*0.05 = 2.4 and the variance is σ² = np(1-p) = 2.4*0.95 = 2.28, hence its standard deviation is σ = √2.28 = 1.51.

Note that, if approximate probability is asked, we could just approximate X with a Normal random variable with mean 2.4 and standard deviation 1.51 (this can be done because of the central limit theorem). We will calculate the probability manually. He wont foul out if he picks 0,1,2,3 or 4 fouls, thus

$P(X \leq 4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \\= 0.95^{48}+48*0.95^{47}*0.05+{48 \choose 2}*0.95^{46}*0.05^2+\\{48 \choose 3}*0.95^{45}*0.05^3+{48 \choose 4}*0.95^{44}*0.05^4 \\= 0.085+0.215+0.266+0.215+0.127 = 0.9093$