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kipiarov [429]
3 years ago
7

A child is 2 -1/2 feet tall. The child’s mother is twice as tall as the child. How tall is the child’s mother

Mathematics
2 answers:
Len [333]3 years ago
7 0

Answer:

  5 feet

Step-by-step explanation:

"Twice as tall" means "2 times as tall".

  2 × (2 1/2 ft) = (2 × 2 ft) +(2 × (1/2 ft)) = 4 ft + 1 ft = 5 ft

The child's mother is 5 feet tall.

bagirrra123 [75]3 years ago
4 0

Answer:

The mother is 5ft tall

Step-by-step explanation:

2 1/2 + 2 1/2 = 5ft

2ft+2ft = 4ft

1/2+1/2= 1ft

4ft+1ft = 5ft

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What is 52.0 divided by 21
kicyunya [14]

Going through long division:  We see that 21 goes into 52 twice, with a remainder of 10.  Bringing down the first 0 after '21,' we have '100' to divide by 21; the quotient is 4 and the remainder 16.  Bringing down the 2nd '0' after 21, we now have '160' to divide by 21; the quotient is 7 with a remainder of 13.  So far we have 2.47 as our quotient.  We can continue this process as far as desired.  

Divide 21 into 52 with a calculator to check your work here.

52 divided by 21 is 2.48 after rounding up.

8 0
4 years ago
Need help asap please
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Answer:

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Step-by-step explanation:

3 0
3 years ago
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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%.

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4 years ago
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Greatest Common Factor: 2

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3 years ago
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