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saw5 [17]
3 years ago
6

Justin did push-ups every weekday this week. He did 888 push-ups on Monday, 14 push-ups on Tuesday, 18 push-ups on Wednesday, 6

push-ups on Thursday, and 8 push-ups on Friday.
Find the mean number of push-ups.
Mathematics
1 answer:
Svetlanka [38]3 years ago
7 0

Answer:

8

Step-by-step explanatio

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Find the difference between the actual and estimated value of 55-21​
Airida [17]

Answer:

55-21=34

50-20=30

hope it helped :)

5 0
3 years ago
Please help! I have this question and one more I'm about to post, if you can answer them that would be most appreciated. These a
dimaraw [331]
D is the answer to this problem
4 0
3 years ago
Please helppp (had to have 20 words haha)
s2008m [1.1K]

Answer:

w==77

Step-by-step explanation:

The sum of the 2 opposite interior angles equals the exterior angle.

w-36 +w-20=w+21

combine like terms: 2w-56=w+21

subtract w from both sides: w-56=21

add 56 to both sides: <u>w=77</u>

6 0
4 years ago
2,803 x 406. Estimate the product first. ≈ ________________ × ________________
Advocard [28]

Answer:

approximately 1,148,000

Step-by-step explanation:Round 2,803 to 2,800 and 406 to 410 then multiply 2,800 by 410

8 0
3 years ago
assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the
drek231 [11]

Answer:

65.3658 inches

Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value x_{0} such that P(X < x_{0}) = 0.76, but, 0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5), i.e., (x_{0}-63.6)/2.5 is the 76th percentile of the standard normal distribution. So, (x_{0}-63.6)/2.5 = 0.7063, x_{0} =63.6+(2.5)(0.7063) = 65.3658. Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.

7 0
4 years ago
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