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2 years ago

Im almost done with it

Mathematics

1 answer:

baherus [9]2 years ago

4 0

Answer:

Area of rectangle = $\mathbf{x^2-4x}$

Perimeter of rectangle = $\mathbf{4x-8}$

Step-by-step explanation:

We are given:

Length of rectangle = x

Width of rectangle = x-4

We are not given if we want to find area of rectangle or perimeter of rectangle.

So, I will be finding both

Area of rectangle

The formula used is: $Area\: of\: rectangle=Length \times Width$

Putting values and finding area

$Area\: of\: rectangle=Length \times Width\\Area\: of\: rectangle=x \times (x-4)\\Area\: of\: rectangle=x^2-4x$

So, we get area of rectangle = $\mathbf{x^2-4x}$

Perimeter of rectangle

The formula used is: $Perimeter \: of\: rectangle=2(Length+ Width)$

Putting values and finding perimeter

$Perimeter \: of\: rectangle=2(Length+ Width)\\Perimeter \: of\: rectangle=2(x+ x-4)\\Perimeter \: of\: rectangle=2(2x-4)\\Perimeter \: of\: rectangle=4x-8$

So, we get perimeter of rectangle = $\mathbf{4x-8}$

I hope, it can help in solving the question.

You might be interested in

a bag contains 8 red balls and 6 green balls. michelle randomly draws 3 balls from the bag. she does not replace the balls after

earnstyle [38]

Answer:

Red balls 2/14

Green balls 1/14

Step-by-step explanation:

I hope that helped.

6 0

2 years ago

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

3 years ago

Read 2 more answers

Help me please!..

AlexFokin [52]

Answer:

44 times

Step-by-step explanation:

There is a 50% (2/4) chance that this could occur, as given in the question.

88*.5 = 44.

8 0

2 years ago