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kodGreya [7K]
2 years ago
5

(7th grade work) which is the best estimate for 0.8% of 503? Choices: 10 4 1

Mathematics
2 answers:
elixir [45]2 years ago
7 0

Answer:

I think it's 4.

Step-by-step explanation:

Andru [333]2 years ago
5 0
ANSWER:
It’s 4 it’s the best choice
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It is J

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Ray CE is the angle bisector of ACD. Which statement about the figure must be true?
Viefleur [7K]

we know that

"Bisect" means to divide into two equal parts

so

If the ray CE is the angle bisector of ADC

then

Angle(ACD)=Angle(ACE)+Angle(ECD)

Angle(ACE)=Angle(ECD)

so

Angle(ACD)=2Angle(ACE)

Angle(ACE)=\frac{1}{2} Angle(ACD)

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<u>the answer is the option</u>

Angle(ACE)=\frac{1}{2} Angle(ACD)

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7 0
2 years ago
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How do you solve ? <br>2(x−(3+2x)+9)=3x−8
PIT_PIT [208]

Hey there!!

Given equation :

... 2 ( x - ( 3 + 2x ) + 9 ) = 3x - 8

Using the distributive property.

... 2 ( x - 3 - 2x + 9 ) = 3x - 8

... 2 ( -x + 6 ) = 3x - 8

Using the distributive property.

... -2x + 12 = 3x - 8

Subtracting 12 on both sides.

... -2x = 3x - 8 - 12

... -2x = 3x - 20

Subtracting 3x on both sides.

... -2x - 3x = -20

... -5x = -20

Dividing by -5 on both sides.

... x = -20 / -5

... x = 4

<em>Hence, the answer is 4. </em>

Hope my answer helps!

6 0
2 years ago
A baseball player hit 65 home runs in a season. Of the 65 home​ runs, 19 went to right​ field, 23 went to right-center​ field, 9
olchik [2.2K]

Answer:

a) 29.23% probability that a randomly selected home run was hit to right field

b) 29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes. It is said to be unusual if it is lower than 5% or higher than 95%.

(a) What is the probability that a randomly selected home run was hit to right field?

Desired outcomes:

19 home runs hit to right field

Total outcomes:

65 home runs

19/65 = 0.2923

29.23% probability that a randomly selected home run was hit to right field

(b) Was it unusual for this player to hit a home run to right field?

29.23% probability that a randomly selected home run was hit to right field, which is not lower than 5% nor it is higher than 95%. So it was not unusual for this player to hit a home run to right field.

8 0
3 years ago
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