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malfutka [58]
3 years ago
11

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated b

y a normal distribution with a mean value of 150 sec and a standard deviation of 15 sec. The fastest 10% are to be given advanced training. What task times qualify individuals for such training? (Round the answer to one decimal place.)
Mathematics
1 answer:
zepelin [54]3 years ago
8 0

Answer:

If the task is performed in less than or equal to 130.8 seconds, then, the individuals qualify for advanced training.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 sec

Standard Deviation, σ = 15 sec

We are given that the distribution of time taken is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.10

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 150}{15})=0.10  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 150}{15} = -1.282\\x = 130.8  

Thus, if the task is performed in less than or equal to 130.8 seconds, then, the individuals qualify for advanced training.

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3 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
Which of the following are progressions?
professor190 [17]

In mathematics: Arithmetic progression, sequence of numbers such that the difference of any two successive members of the sequence is a constant. Geometric progression, sequence of numbers such that the quotient of any two successive members of the sequence is a constant. so i chose b)Sequences

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In order for figure EFGH to be a trapozoid, which of the following must be true?
kicyunya [14]
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3 years ago
beni is shopping for a new cell phone plan. company a charges$72.25for a new phone and $85.50 per month.company b charges $151.2
tigry1 [53]
Missing questions and subsequent solutions:
(a) Write an equation for company A for cost, C, number of months, n, that Beni will pay for the phone.

Solution:
For company A:
C = 72.25 + 85.50n

(b) Write an eqyation for company B for cost, C, and number of months, n, that Bei will pay for the phone.

Solution:
For company B:
C = 151.25 + 65.75n

(c) Write an inequality when the cost from company A is better than cost from company B.

Solution:
72.25 + 85.50n ≤ 151.25 + 65.75n
(85.50-65.75)n ≤ (151.25 - 72.25)
19.75 n ≤ 79
n ≤ 4

(d) Value of n for which cost from the two companies will be the same.

Solution:
If cost for companies A and B are the same, then
72.25 + 85.50n = 151.25 + 65.75n
(85.5 - 65.75)n = 151.25 - 72.25
19.75n = 79
 n = 79/19.75 = 4 months

After 4 months,
C = 72.25 + 85.5*4 = $414.25
7 0
3 years ago
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