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galina1969 [7]
2 years ago
13

Pregnancy length (in days) has a normal probability distribution with a mean of 266 days and a standard deviation of 16 days. Wh

at is the length of the pregnancy that marks the start of the 25th percentile? Enter a number rounded to two decimal places. Do not enter the units.
Mathematics
1 answer:
Oxana [17]2 years ago
4 0

Answer:

264.20

Step-by-step explanation:

Let X be the pregnancy length (in days)

Then X is N(266,16)

We can convert X to Z by

\frac{x-266}{16}

Z=\frac{x-266}{16} is N(0,1)

25th percentile for Z is

-0.675

Convert back to x score

x=-0.675(16)+275\\X =264.20

25th percentile is 264.20

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Which of the following represents the zeros of f(x) = 5x3 − 6x2 − 59x + 12? (2 points)
Furkat [3]

Answer:

4, −3, 1 over 5

Step-by-step explanation:

x is a zero of f(x) if f(x) = 0

In this problem, we have that:

f(x) = 5x^{3} - 6x^{2} - 59x + 12

4, 3, 1 over 5

f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0

So 4 is a zero of the function

f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84

So 3 is not a zero of the function, and this option is incorrect

4, 3, − 1 over 5

f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84

So 3 is not a zero of the function, and this option is incorrect

4, −3, 1 over 5

f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0

So 4 is a zero of the function

f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0

So -3 is a zero of the function

f(\frac{1}{5}) = f(0.2) = 5*(0.2)^{3} - 6*(0.2)^{2} - 59*(0.2) + 12 = 0

So 1 over 5 is a zero of the function

This is the correct answer.

4, −3, −1 over 5

f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0

So 4 is a zero of the function

f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0

So -3 is a zero of the function

f(-\frac{1}{5}) = f(-0.2) = 5*(-0.2)^{3} - 6*(-0.2)^{2} - 59*(-0.2) + 12 = 23.52

-1 over 5 is not a zero of the function

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2 years ago
A theater company is considering raising the price of its tickets. It currently charges $8.50 for each ticket and sells an avera
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Answer:

The answer is c.

Step-by-step explanation:

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2 years ago
In a particular class, the probability of a student having blue eyes is 0.3, of having both blue eyes and blonde hair is 0.2, an
kramer
Let the blue-eyed student set is B Let the blonde student set is C<span>
P(B)=0.3</span> <span>
P(BC)=0.2</span> <span>
P(B′C′)=0.5

Therefore, the answer is B. 0.2.

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2 years ago
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The Brutus Gourmet Company produces delicious organic dog treats for canines with discriminating tastes. Management wants the bo
KiRa [710]

Answer:

Average range(R-bar) = Sum of the sample range for each sample/number of samples = (9+8+1+8+7) /5 = 33/5 = 6.6

Sample size = 8

For a sample size of 8 the factors for control limit for range (D3) = 0.136 (obtained from table of constraints for x-bar and R chart

LCL = R-bar × D3 = 6.6 × 0.136 = 0.90

Step-by-step explanation:

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The Yearbook Club is going to an amusement park, and each of their 12 members will pay for admission and will also help pay for
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Answer:

i honestly dont know bro i tried a lot of ways but nothing was equivalent

Step-by-step explanation:

3 0
2 years ago
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