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lbvjy [14]
3 years ago
5

a. A 2017 poll found that 56​% of college students were very confident that their major will lead to a good job. If 15 college s

tudents are chosen at​ random, what's the probability that 13 of them were very confident their major would lead to a good​ job? Let a success be a college student being very confident their major would lead to a good job.
Mathematics
1 answer:
lesya [120]3 years ago
8 0

Answer:

0.01083 or 1.083%

Step-by-step explanation:

This problem can be modeled as a binomial probability model with probability of success p = 0.56.

The probability of x=13 successes (a college student being very confident their major would lead to a good job) in a number of trials of n=15 is:

P(X=x) = \frac{n!}{(n-x)!x!} *p^x*(1-p)^{n-x}\\P(X=13) = \frac{15!}{(15-13)!13!} *0.56^{13}*(1-0.56)^{15-13}\\P(X=13) = 0.01083=1.083\%

The probability is 0.01083 or 1.083%.

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Given the graphs of f(x) =x -7 and g(x) =-5x-1, what is the solution to the equation f() = g(x)?
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Answer:

1

Step-by-step explanation:

x-7=-5x-1

x-(-5x)-7=-1

x+5x-7=-1

6x-7=-1

6x=-1+7

6x=6

x=6/6

x=1

8 0
3 years ago
The measure of the base of a triangle is represented by 4x+3 and the height is 6x represent the area of the triangle as a polyno
Juliette [100K]

Answer:

12x^2+9x

Step-by-step explanation:

The formula of the area of a triangle is \frac{bh}{2}. Plugging in the given base and height values, you get \frac{(4x+3)(6x)}{2}=\frac{24x^2+18x}{2}=12x^2+9x. Hope this helps!

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HEY GUYS THIS IS MY FINAL MATH ASSIGNMENT OF THE YEAR... PLS JUST HELP ME FIND THE DOMAIN AND RANGE THANK YOU
Gnom [1K]

Answer:

domain: all real numbers (-∞,∞)

range: y ≤ 64

Step-by-step explanation:

since the domain is arn,

and range is the coordinates

coordniate (top or bottom-most point)

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(2, 64)

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3 0
2 years ago
Read 2 more answers
Witch is grater 3/4 or 11/16 ??
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<h3>Hey there! </h3><h3>\frac{3}{4} → 0.75 → 0.75%</h3><h3>\frac{11}{16} → 0.6875 → 0.68% </h3><h3>So, the one that is GREATER is: \frac{3}{4} because 75% is bigger than</h3><h3> 68%</h3><h3>Good luck on your assignment and enjoy your day! </h3><h3>~LoveYourselfFirst:) </h3>
5 0
3 years ago
sample of 150 Publix cashiers showed a mean hourly wage of $ 8.00, with a standard deviation of $ 1. Give an interval that is li
Scorpion4ik [409]

Answer:

B. (6, 10)

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 8

Standard deviation = 1

Give an interval that is likely to contain about 95% of the sampled cashiers' hourly wages.

By the Empirical Rule, 95% of the sampled cashiers' hourly wages will be within 2 standard deviations of the mean, so from 2 standard deviations below the mean to two standard deviations above the mean

Two standard deviations below the mean:

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Two standard deviations above the mean

8 + 2*1 = 10

So the correct answer is:

B. (6, 10)

4 0
3 years ago
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