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3 years ago

What is the area of parallelogram if base is 13 and the height 14

Mathematics

1 answer:

Marianna [84]3 years ago

6 0

Answer:

A = 182

Step-by-step explanation:

14 x 13 = 182

So, 182 would be your answer.

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.) What is the Greatest Common Factor of x5,x4, and x3?

Viefleur [7K]

40 common factors multiplication

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3 years ago

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Ashley has a rectangle made out of paper that is 8 cm by12 cm. She folds it in half twice, first vertically and then horizontall

Svetach [21]

Answer:

$Area =24cm^2$

Step-by-step explanation:

Given

$L = 8cm$

$W = 12cm$

$r = 2$ -- folded twice

Required

The area of the new rectangle

When the length was folder, the new length is:

$l = L/2 = 8cm/2 = 4cm$

When the width was folder, the new width is:

$w = W/2 = 12cm/2 = 6cm$

So, the new area is:

$Area =l * w$

$Area =4cm * 6cm$

$Area =24cm^2$

8 0

2 years ago

11] (8•10^-3)(2•10^10

Oduvanchick [21]

Answer:

16*10^7 or 160000000

Step-by-step explanation:

combine whole numbers: 8*2

and combine exponents: 10^-3*10^10=10^7

8*2*10^7

16*10^7

160000000

5 0

3 years ago

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$