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ANEK [815]
2 years ago
8

. A manager has just received the expense checks for six of her employees. She randomly distributes the checks to the six employ

ees. What is the probability that exactly five of them will receive the correct checks (checks with the correct names)?
Mathematics
1 answer:
Tcecarenko [31]2 years ago
3 0

Answer:

13,8%

Step-by-step explanation:

There are six employees and six cheks, so there are 36 (6x6) possible combinations so if we need to measure the probability that five of them receive the exact check  is only one for each one of them over the 36 possibilities, so 1/36 for one plus 1/36 the second and so on.  5/36 = 13,8%.

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Juanita rolls a number cube x times. For which value of x are the experimental probability and the theoretical probability most
vekshin1

Answer:

1,000

Step-by-step explanation:

The more rolls you make, the closer the experimental and theoretical probabilities get closer together

6 0
2 years ago
A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was co
Firlakuza [10]

Answer:

Step-by-step explanation:

1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88

The correct option is D

2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p

p = 70/100 = 0.7

q = 1 - p = 1 - 0.7 = 0.3

n = number of selected users = 50

Mean, u = np = 50×0.7 = 35

Standard deviation, u = √npq = √50×0.7×0.3 = 3.24

x = number of internet users

The formula for normal distribution is expressed as

z = (x - u)/s

We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as

P(x lesser than or equal to 25)

The z value will be

z = (25- 35)/3.24 = - 10/3.24 = -3.09

Looking at the normal distribution table, the corresponding z score is 0.001

P(x lesser than or equal to 25) = 0.001

7 0
2 years ago
A simple random sample of 100 concert tickets was drawn from a normal population. The mean and standard deviation of the sample
alexandr402 [8]

Answer:

We accept the null hypothesis and the population mean is $120.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100

Sample mean, \bar{x} = $120

Alpha, α = 0.01

Sample standard deviation, s = $25

First, we design the null and the alternate hypothesis

H_{0}: \mu = 125\\H_A: \mu \neq 125

We use two-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = -2.0              

p-value one tail= 0.024

p-value two tail= 0.048

Conclusion:

Since the p-value for two tailed test is greater than the significance level, we fail to reject the null hypothesis and accept it.

Thus, the population mean is $120.

6 0
2 years ago
The green turtle lays eggs that are approximately spherical with an average diameter of 4.5 centimeters. each turtle lays an ave
Art [367]
The average volume if these eggs will be found as follows:
Volume of one egg is:
V=4/3πr²
V=4/3×π×(4.5/2)^3
V=71.57 cm³
Given that the turtle lays on average 113 eggs, thus the total volume of eggs will be:
(113×71.57)
=8187.41~8187 cm³

4 0
1 year ago
Show that the following rational number by expressing it as a ratio quotient of two integers 0.9
kipiarov [429]
A rational number can be expressed in the form a/b, where a and b are other integers.  To satisfy this definition, 0.9 can be written as 9/10, 18/20, 90/100, etc.
7 0
2 years ago
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