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Ray Of Light [21]
3 years ago
15

Adult male heights have a normal probability distribution with a mean of 70 inches and a standard deviation of 4 inches.

Mathematics
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer: 0.1587

Step-by-step explanation:

Given : Adult male heights have a normal probability distribution with a mean of 70 inches and a standard deviation of 4 inches.

i.e. \mu=70 ,\ \sigma=4

Let x be the random variable that represents the heights of adult males.

Also, z=\dfrac{x-\mu}{\sigma}

For x= 74 , we have

z=\dfrac{74-70}{4}=1

Then, the probability that a randomly selected male is more than 74 inches tall:-

P(x>74)=P(z>1)=1-P(z\leq1)\\\\=1- 0.8413447\\\\=0.1586553\\\\\approx0.1587 [using z-value table.]

Hence, the probability that a randomly selected male is more than 74 inches tall = 0.1587

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3 years ago
Which multiplication expression is equivalent to<br>6+ (6) + 16)​
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2 years ago
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dusya [7]

Answer:

The sum is

7.92\times 10^3

Step-by-step explanation:

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6 0
3 years ago
There are 8 tennis players in a round robin tournament, which means each team will play every other team once. How many matches
Aliun [14]
<h3>Answer:  28</h3>

==========================================================

Explanation:

Method 1

Imagine a table with 8 rows and 8 columns to represent all possible match-ups. You can actually draw out this table or just think of it as a thought experiment.

There are 8*8 = 64 entries in the table. Along the northwest diagonal, we have each team pair up with itself. This is of course silly and impossible. We cross off this entire diagonal so we drop to 64-8 = 56 entries.

Then notice that the lower left corner is a mirror copy of the upper right corner. A match-up like AB is the same as BA. So we must divide by 2 to get 56/2 = 28 different matches.

----------------------------------

Method 2

There are 8 selections for the first slot, and 8-1 = 7 selections for the second slot. We have 8*7 = 56 permutations and 56/2 = 28 combinations.

----------------------------------

Method 3

Use the nCr combination formula with n = 8 and r = 2

n C r = \frac{n!}{r!(n-r)!}\\\\8 C 2 = \frac{8!}{2!*(8-2)!}\\\\8 C 2 = \frac{8!}{2!*6!}\\\\8 C 2 = \frac{8*7*6!}{2!*6!}\\\\ 8 C 2 = \frac{8*7}{2!}\\\\ 8 C 2 = \frac{8*7}{2*1}\\\\ 8 C 2 = \frac{56}{2}\\\\ 8 C 2 = 28\\\\

There are 28 combinations possible. Order doesn't matter (eg: match-up AB is the same as match-up BA).

Notice how the (8*7)/2 expression is part of the steps shown above in the nCr formula.

7 0
1 year ago
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Answer:

 What we are going to do for this case is find the cost for each case:

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Read more on Brainly.com - brainly.com/question/2528459#readmoreStep-by-step explanation:

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