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Ad libitum [116K]

3 years ago

6

What is the answer for 7.42 ÷ 1.75 = ______

2 answers:

Colt1911 [192]3 years ago

6 0

Answer:

4.24

Step-by-step explanation:

IRINA_888 [86]3 years ago

6 0

The answer is 4.24

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jejsjakaokwnajnakwnaknajqnwkanwjbaka (sorry i need 20 characters)

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if a rectangle has an area of 30 square meters and a perimeter of 34 meters . what is the dimensions of the rectangle?

timama [110]

P=2(L+W) and P=34 so

2(L+W)=34

L+W=17 so we can say

L=17-W

A=LW using L from above

A=(17-W)W

A=17W-W^2 and A=30 so

30=17W-W^2

W^2-17W+30=0

W^2-2W-15W+30=0

W(W-2)-15(W-2)=0

(W-15)(W-2)=0

So the dimensions of the rectangle are 15 meters by 2 meters.

7 0

3 years ago

Which number is composite?<br> 31<br> 43<br> 37<br> 49

Alina [70]

A composite number is one that can be made by multiply two numbers to each other. Since 49=7*7, 49 is a composite number.

Check:

You can check a multiplication chart, but you will see that 49 is the only one that can be multiplied by 2 numbers

4 0

3 years ago

Yemi saves 150 daily how much will it save in two weeks

Leokris [45]

Answer:

$2,100

Step-by-step explanation:

In two weeks there is 14 days so, I times 150 by 14 which is 2100.

3 0

3 years ago

Read 2 more answers

A company negotiator claims that only 35% of union members will support a strike, but a union rep believes that the true percent

Softa [21]

Answer:

$(d)$ $H_o: p = 0.35$ $H_a: p > 0.35$

Step-by-step explanation:

Given

$\mu = 35\%$

Required

The hypothesis statements

First claim: 35% will support a strike.

This represents the null hypothesis

$H_o: p = 0.35$

Second claim: A greater percentage will support the strike.

This represents the alternate hypothesis

$H_a: p > 0.35$

3 0

3 years ago

A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla

nekit [7.7K]

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = $P(A)= \frac{13}{100}=0.13$

The probability that athlete is a basketball player = $P(B)= \frac{52}{100}=0.52$

The probability that athlete is both basket ball player and football player = $P(A\cap B) = \frac{9}{100}=0.09$

We have to find the probability that an athlete chosen is either a football player or a basketball player $P(A\cup B)$ .

Now we know that;

$P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%$

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

5 0

3 years ago