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vekshin1
3 years ago
7

Question 2

Mathematics
2 answers:
Zinaida [17]3 years ago
7 0
3 and 1/2
The polynomial would be ^^
timurjin [86]3 years ago
4 0

There are two zero of the polynomial...The answers are 3 and 1/2......

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A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 623 babies bo
Margaret [11]

Answer:

The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 3448, \sigma = 862

Proportion of newborns who weighed between 1724 grams and 5172 grams.

This is the pvalue of Z when X = 5172 subtracted by the pvalue of Z when X = 1724. So

X = 5172

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{5172 - 3448}{862}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 1724

Z = \frac{X - \mu}{s}

Z = \frac{1724 - 3448}{862}

Z = -2

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

Estimate the number of newborns who weighed between 1724 grams and 5172 grams.

0.9544 out of 623 babies. SO

0.9544*623 = 595

The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.

5 0
3 years ago
a cone shaped container has a height of 9 inches and a diameter or 2in it is filled with a liquid that is worth $2 per cubic inc
Mice21 [21]

Answer:

$18.84

Step-by-step explanation:

The area of a cone is (1/3)pi*r^2*h

With all variables plugged in we get an area of (1/3)*3.14*(1^2)*9, or 9.42 cubic inches. With a value of 2$/in^3, we get 2*9.42, or 18.84.

Therefore, the value of the container is $18.84.

5 0
2 years ago
Need help on this as soon as possible
Andrew [12]
Hypotenuse of the right triangle: h=sqrt(4^2+6^2)
h=sqrt(16+36)
h=sqrt(52)
h=sqrt(4*13)
h=sqrt(4) sqrt(13)
h=2 sqrt(13)

Total Area: T.A.=2*(4*6/2)+[4+6+2 sqrt(13)] (8)
T.A.=2*(12)+[10+2 sqrt(13)](8)
T.A.=24+(10)(8)+[2 sqrt(13)](8)
T.A.=24+80+16 sqrt(13)
T.A.=104+16 sqrt(13)

Answer: 
( 104+16 sqrt(13) )
5 0
3 years ago
Explain how the Remainder Theorem and the Factor Theorem are related.​
Daniel [21]
Basically the remainder theorem links the remainder of division by a binomial with the value of a function at a point while the factor theorem links the factors of polynomial to its zeros
8 0
2 years ago
PLEASE HELP THIS IS THE LAST QUESTION AND I WANT TO GO TO BED. PLEASE EXPLAIN SO I CAN DO IT FOR NEXT TIME AS WELL
rewona [7]
15+70+45=130 :)) that’s how u calculate the perimeter just add them all together.
7 0
3 years ago
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