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Katarina [22]
3 years ago
10

At which price does quantity demanded equal quantity supplied?

Mathematics
1 answer:
yulyashka [42]3 years ago
3 0

the price of the object

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Answer The Factorial (17-12)
Kamila [148]
(17-12) is equal to 5
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3 years ago
If 8/9 of a number is 328, find the number
Gre4nikov [31]

Answer: the number is 369

Step-by-step explanation:

Let the missing number be y

8/9 × y = 328

8/9y = 328

Divide both sides by 8/9

y = 328 ÷ 8/9

y = 328 × 9/8 (multiply the two numerators: 328 x 9= 2952

Multiply the denominator by 1: 8 x 1= 8

Therefore y = 2952/8

y = 369

I hope this helps.

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3 years ago
What is the GCF of these two
kkurt [141]

Answer:

B!!!

Step-by-step explanation:

6 0
3 years ago
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Solve for c 3c − 10 = 2c
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Hopefully this helps
C=10
6 0
3 years ago
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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
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