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3 years ago

Simplify the expression by combining like terms.

Mathematics

1 answer:

klasskru [66]3 years ago

3 0

My guess would probably be A

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There are two spinners containing only red and orange slices. Spinner A has 7 red slices and 13 orange slices. All the slices ar

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Event 4 (impossible), Event 3 (probability is 7/20), Event 1 (probability is 6/12 which is greater than 7/20), and Event 2 (certain to happen)

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3 years ago

Simplify (4.5)(5)(−2). (1 point) 45 4.5 −4.5 −45

nikitadnepr [17]

The answer is 4.5*5*-2= -45 Thanks!

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3 years ago

A contractor is given a scale drawing of a rectangular patio. The scale from the patio to the drawing is 4ft to 1in. What is the

AnnyKZ [126]

Answer:

170 ft²

Step-by-step explanation:

Given:

Patio drawing of 4.25 in by 2.5 in

Scale of Patio to its drawing = 4ft to 1 in

Requires:

Area of actual Patio

SOLUTION:

Area of actual Patio = area of a rectangle = length × width

Given a scale drawing of 4ft to 1 in, therefore:

Length of actual Patio = 4.25 × 4 = 17 ft

Width of actual Patio = 2.5 × 4 = 10 ft

Therefore:

Area of actual Patio = 17 ft × 10 ft = 170 ft²

5 0

2 years ago

Based on the picture below. find the angle measure for lmn.

trasher [3.6K]

Answer:

Step-by-step explanation:

4x + 2x = 90

6x = 90

x = 15

4(15)= 60 for <LMN

2(15)= 30

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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