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Ivahew [28]
3 years ago
8

Convert 32 feet per second to kilometers per hour. (Hint: There are 5280 feet in a mile, and one mile is 1.61 kilometers)​

Mathematics
2 answers:
iris [78.8K]3 years ago
7 0

Answer: 35.113

Hope this helps!

LUCKY_DIMON [66]3 years ago
7 0

Answer:

Answer: 35.113

Step-by-step explanation:

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1
Blizzard [7]
I believe it is false.
7 0
3 years ago
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On Friday, Oscar sold 2 pitchers of lemon are from his lemonade stand. On Saturday, he sold 1/2 as much lemonade as on Friday. H
Ann [662]

Answer:

The correct answer is 1 pitcher.

Step-by-step explanation:

Oscar sold 2 pitchers of lemon are from his lemonade stand on Friday.

On Saturday, he sold \frac{1}{2} as much lemonade as on Friday.

The number of pitchers of lemon from the lemonade stand which Oscar sold on Saturday is given by \frac{1}{2} × 2 = 1 pitcher.

Thus Oscar sold 1 pitcher on Saturday which is half of the number of pitchers of lemon he could sell on Friday.

4 0
3 years ago
The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0
3 years ago
50 points!!! Will mark brainliest if correct!! Pls help me!!
goldenfox [79]

Answer:

3rd option is correct, give me brainiest

4 0
2 years ago
HELP ASAP
slega [8]

Answer:

No

Step-by-step explanation:

The sample size wasn't big enough. It could only be a trend in that grade.

7 0
2 years ago
Read 2 more answers
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