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Triss [41]
3 years ago
13

I need help Answering this question anyone

Mathematics
1 answer:
Eduardwww [97]3 years ago
3 0
Y=2x-1 is the answer
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What is the solution to 2x+10y=-14 and x+5y=-7 answer using elimination with steps
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The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
inessss [21]

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean \mu =266\\

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find P(X\geq 308)

Using normal distribution,

z=\frac{X-\mu}{\sigma}

z=\frac{308-266}{15}

=\frac{42}{15}

Thus z=2.8

So P\left (X\geq 308  \right )=1-P(X

=1-P(z

=1-Table\:  value\:  of\:  2.8

=1-0.99744

=0.00256

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3​%.

The z-value for the lowest of 3% is -1.8808

Then X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu

This implies X=-1.8808*15+266=237.788

Thus the babies who are born on or before 238 days are considered to be premature.

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3 years ago
What is the multiplicative inverse of 21/2
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Simplify the expression
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I think the answer is 1/m^18
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