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DanielleElmas [232]
3 years ago
13

How to solve this one​

Mathematics
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

2x + y =6

x + 2y = —6

Step-by-step explanation:

I hope that's help

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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.

5 0
3 years ago
Two geometric means are inserted between 6 and 384 so the four numbers form a geometric sequence. What are these two numbers?
8_murik_8 [283]

Answer:

A

Step-by-step explanation:

idc

5 0
3 years ago
Can someone please help me
MA_775_DIABLO [31]

Answer:

19

Step-by-step explanation:

12 2/3 divided by 2/3 = 19

4 0
3 years ago
ANSWER IN THE COMMENTS SECTION
ryzh [129]

Answer:

B.

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
What is 1*1000000*5*6/3*1757/4*45567-4-6-8*5846584*0+1+2+3+4+5+6+7+8+9+10
astra-53 [7]

Answer:

2.0015305e+15 is the answer

5 0
2 years ago
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