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PSYCHO15rus [73]

2 years ago

13

The employees of a firm that manufactures insulation are being tested for indications of asbestos in their lungs. The firm is re

quested to send three employees who have positive indications of asbestos to a medical center for further testing. If 30% of the employees have positive indications of asbestos in their lungs, find the probability that eleven employees must be tested in order to find three positives. (Round your answer to three decimal places.)

1 answer:

diamong [38]2 years ago

8 0

Answer:

0.070

Step-by-step explanation:

Y = number on trial

Y has a negative binomial distribution

r = 3

P = 30% = 0.3 probability of positive indication.

P(Y = 11) probability of 11 employees that must be tested to get 3 positives

Y-1Cr-1*p^r*q^(y-r)

Y-1 = 11-1 = 10

r-1 = 3 -1 = 2

10C2 x 0.3³x0.7⁸

45x0.027x0.05764801

= 0.070

This is the probability that 11 employees must be tested to get 3 positives.

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A ticket to Silver Dollar City costs $125. You have a coupon that will save you 33%. How much will you SAVE? In the same scenari

vampirchik [111]

You will save 41.25 dollars, and you will pay 83.75 dollars if you use the coupon.
To get the percent of money you save you multiply 125 dollars by 33% to get your answer. To get the number of dollars you will take the amount you got by multiplying 125 dollars by 33% and subtract it from 125, so that’s 125 - 41.25 = 83.75

7 0

2 years ago

Which pair of monomials has the least common multiple (LCM) of 54x2y3?

Lera25 [3.4K]

ANSWER

The correct answer is D.

EXPLANATION

If we express the monomial,

$18 {x}^{2} y$

as product of primes, we obtain:

$2 \times {3}^{2} \times {x}^{2}y$

If we express the monomial

$27x {y}^{3}$

as product of primes we obtain:

$= {3}^{3} \times x {y}^{3}$

The least common multiple of these two binomials is the product of the highest powers of the common factors.

The LCM is

$= 2 \times {3}^{3} \times {x}^{2} {y}^{3}$

$=54 {x}^{2} {y}^{3}$

Therefore the correct answer is D.

6 0

2 years ago

Read 2 more answers

Simplify 150 -(0.4 x 150)+15

Rufina [12.5K]

150-(0.4 x 150)+15= 105

5 0

3 years ago

Suppose that demand in period 1 was 7 units and the demand in period 2 was 9 units. Assume that the forecast for period 1 was fo

Zepler [3.9K]

Answer:

Step-by-step explanation:

Forecast for period 1 is 5

Demand For Period 1 is 7

Demand for Period 2 is 9

Forecast can be given by

$F_{t+1}=F_t+\alpha (D_t-F_t)$

where

$F_{t+1}=Future Forecast$

$F_t=Present\ Period\ Forecast$

$D_t=Present\ Period\ Demand$

$\alpha =smoothing\ constant$

$F_{t+1}=5+0.2(7-5)$

$F_{t+1}=5.4$

Forecast for Period 3

$F_{t+2}=F_{t+1}+\alpha (D_{t+1}-F_{t+1})$

$F_{t+2}=5.4+0.2\cdot (9-5.4)$

$F_{t+2}=6.12$

8 0

3 years ago

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is

marysya [2.9K]

Answer:

(a) 0.14%

(b) 2.28%

(c) 48%

(d) 68%

(e) 34%

(f) 50%

Step-by-step explanation:

Let <em>X</em> be a random variable representing the prices paid for a particular model of HD television.

It is provided that <em>X</em> follows a normal distribution with mean, <em>μ</em> = $1600 and standard deviation, <em>σ</em> = $100.

(a)

Compute the probability of buyers who paid more than $1900 as follows:

$P(X>1900)=P(\frac{X-\mu}{\sigma}>\frac{1900-1600}{100})$

$=P(Z>3)\\=1-P(Z$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.

(b)

Compute the probability of buyers who paid less than $1400 as follows:

$P(X$

$=P(Z$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.

(c)

Compute the probability of buyers who paid between $1400 and $1600 as follows:

$P(1400$

$=P(-2$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.

(d)

Compute the probability of buyers who paid between $1500 and $1700 as follows:

$P(1500$

$=P(-1$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.

(e)

Compute the probability of buyers who paid between $1600 and $1700 as follows:

$P(1600$

$=P(0$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.

(f)

Compute the probability of buyers who paid between $1600 and $1900 as follows:

$P(1600$

$=P(0$

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.

8 0

2 years ago