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joja [24]
2 years ago
6

A supervisor finds the mean number of miles that the employees in a department live from work. Which mileage is within a z-score

of 1.5? A. 21 miles B. 24 miles C. 36 miles D. 41 miles
Mathematics
2 answers:
Vikentia [17]2 years ago
7 0
The z-score tells you how many standard deviations from the mean. 

<span>1.5 * 3.6 = 5.4 miles </span>

<span>So anything within 5.4 miles of the average (29). </span>

<span>The range would be: </span>
<span>29 - 5.4 = 23.6 </span>
<span>to: </span>
<span>29 + 5.4 = 34.4 </span>

<span>23.6 ≤ x ≤ 34.4 </span>

<span>Answer: </span>
<span>B) 24 miles</span>
AleksandrR [38]2 years ago
6 0

Answer:

24 miles

Step-by-step explanation:

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d1i1m1o1n [39]
Area of a Triangle = 0.5 x Base x Height 
 
Therefore the area of one mosaic triangle = 0.5 x 3 x 5 = 7.5

So the area of the whole mosaic is : 7.5 x 200 = 1500cm^2 
8 0
2 years ago
Which expression is equivalent to (x^6y^8)^3\x^2y^2
Damm [24]

Answer:

\large\boxed{\dfrac{(x^6y^8)^3}{x^2y^2}=x^{16}y^{22}}

Step-by-step explanation:

\dfrac{(x^6y^8)^3}{x^2y^2}\qquad\text{use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\=\dfrac{(x^6)^3(y^8)^3}{x^2y^2}=\dfrac{x^{(6)(3)}y^{(8)(3)}}{x^2y^2}=\dfrac{x^{18}y^{24}}{x^2y^2}\qquad\text{use}\ \dfrac{a^m}{a^n}=a^{m-n}\\\\=x^{18-2}y^{24-2}=x^{16}y^{22}

5 0
2 years ago
Read 2 more answers
The measure of central angle XYZ is StartFraction 3 pi Over 4 EndFraction radians. What is the area of the shaded sector? 32Pi u
Tems11 [23]

Answer:

<em>96π units²</em>

Step-by-step explanation:

Find the diagram attached

Area of a sector is expressed as;

Area of a sector = θ/2π * πr²

Given

θ = 3π/4

r = 16

Substitute into the formula

area of the sector = (3π/4)/2π * π(16)²

area of the sector = 3π/8π * 256π

area of the sector = 3/8 * 256π

area of the sector = 3 * 32π

<em>area of the sector =96π units²</em>

5 0
2 years ago
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According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m
tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

P(A) = 0.8x + 0.25*0.2x = 0.85x

Probability of developing diabetes while being very active:

0.25x of 0.2. So

P(A \cap B) = 0.25x*0.2 = 0.05x

What is the probability that a middle-aged man with diabetes is very active?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0
1 year ago
The mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of
grandymaker [24]

Answer:

At the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Step-by-step explanation:

We are given that the mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of $2,000.

The ship building association wishes to find out whether their welders earn more or less than $20,000 annually.

<u><em>Let </em></u>\mu<u><em> = mean gross annual incomes of certified welders</em></u>

So, Null Hypothesis, H_0 : \mu = $20,000    

Alternate hypothesis, H_A : \mu \neq $20,000

Here, null hypothesis states that the mean income of welders is equal to $20,000.

On the other hand, alternate hypothesis states that the mean income of welders is not $20,000.

Also, the test statistics that would be used here is <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

                              T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean income

            \sigma = population standard deviation = $2,000

            n = sample size

Now, at the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

3 0
2 years ago
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