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Kobotan [32]
2 years ago
8

What's the value of X?​

Mathematics
1 answer:
Sindrei [870]2 years ago
6 0

Answer:

x=55

Step-by-step explanation:

since it’s a right angle do 90*-35*=55*
Hope this helps! ;-)

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The overhead reach distances of adult females are normally distributed with a mean of 205.5 and a standard deviation of 8.6 . a.
Anastasy [175]

Answer:

a) 0.073044

b) 0.75033

c) The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).

Step-by-step explanation:

The overhead reach distances of adult females are normally distributed with a mean of 205.5 and a standard deviation of 8.6 .

a. Find the probability that an individual distance is greater than 218.00 cm.

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 218

μ is the population mean = 205.5

σ is the population standard deviation = 8.6

For x > 218

z = 218 - 205.5/8.6

z = 1.45349

Probability value from Z-Table:

P(x<218) = 0.92696

P(x>218) = 1 - P(x<218) = 0.073044

b. Find the probability that the mean for 15 randomly selected distances is greater than 204.00

When = random number of samples is given, we solve using this z score formula

z = (x-μ)/σ/√n

where

x is the raw score = 204

μ is the population mean = 205.5

σ is the population standard deviation = 8.6

n = 15

For x > 204

Hence

z = 204 - 205.5/8.6/√15

z = -0.67552

Probability value from Z-Table:

P(x<204) = 0.24967

P(x>204) = 1 - P(x<204) = 0.75033

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).

3 0
3 years ago
A coin is flipped 10 times where each flip comes up either heads or tails. How many possible outcomes (a) contain exactly two he
Tems11 [23]

Answer:

a. 45

b. 176

c. 252

Step-by-step explanation:

First take into account the concept of combination and permutation:

In the permutation the order is important and it is signed as follows:

P (n, r) = n! / (n - r)!

In the combination the order is NOT important and is signed as follows:

C (n, r) = n! / r! (n - r)!

Now, to start with part a, which corresponds to a combination because the order here is not important. Thus

 n = 10

r = 2

C (10, 2) = 10! / 2! * (10-2)! = 10! / (2! * 8!) = 45

There are 45 possible scenarios.

Part b, would also be a combination, defined as follows

n = 10

r <= 3

Therefore, several cases must be made:

C (10, 0) = 10! / 0! * (10-0)! = 10! / (0! * 10!) = 1

C (10, 1) = 10! / 1! * (10-1)! = 10! / (1! * 9!) = 10

C (10, 2) = 10! / 2! * (10-2)! = 10! / (2! * 8!) = 45

C (10, 3) = 10! / 3! * (10-3)! = 10! / (2! * 7!) = 120

The sum of all these scenarios would give us the number of possible total scenarios:

1 + 10 + 45 + 120 = 176 possible total scenarios.

part c, also corresponds to a combination, and to be equal it must be divided by two since the coin is thrown 10 times, it would be 10/2 = 5, that is our r = 5

Knowing this, the combination formula is applied:

C (10, 5) = 10! / 5! * (10-5)! = 10! / (2! * 5!) = 252

252 possible scenarios to be the same amount of heads and tails.

6 0
3 years ago
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