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Neko [114]
3 years ago
9

The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. What is t

he probability that a pregnancy last at least 300​ days? Round to four decimal places. A. 0.0164 B. 0.9834 C. 0.4834 D. 0.0179
Mathematics
1 answer:
AfilCa [17]3 years ago
7 0

Answer: A. 0.0164

Step-by-step explanation:

Given : The lengths of human pregnancies are normally distributed with a mean \mu=268\text{ days}

Standard deviation : \sigma=15\text{ days}

Let X be the random variable that represents the length of pregnancy of a randomly selected human .

z-score : z=\dfrac{X-\mu}{\sigma}

For X = 300

z=\dfrac{300-268}{15}\approx2.13

Now, the probability that a pregnancy last at least 300​ days will be :-

P(X\geq300)=P(z\geq 2.1333)=1-P(z

Hence, the probability that a pregnancy last at least 300​ days =0.0164

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The missing angle is 32°.

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4 years ago
Someone please help me with these too://
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Answer:

1.

The slope is 2

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equation: y = 2x - 10

2.

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y-intercept: 1

Equation: y = 2.4x + 1

Step-by-step explanation:

1.

x      y

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4      16

8      24

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from 12 to 16 you add 4

from 2 to 4 you add 2

y/x = 4/2 = 2

The slope is 2

The y-intercept is -10

equation: y = 2x - 10

I got the y-intercept because I went from 12 to 0 which is -12, and then 2 - 12 and it's -10

__________________________________________________________

2.

x       y

5      4

20   10

30    14

35    16

from 4 to 10, u add 6

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y/x = 15/6 = 2.5

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y-intercept: 1

Equation: y = 2.4x + 1

I got the y-intercept because I went from 4 to 0 which is -4, and 5 - 4 is 1.

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3 years ago
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