What is the GCF of

Mathematics

2 answers:

Troyanec [42]3 years ago

7 0

$1.) 14b^2 - 35b^3=2\cdot7b^2-5b\cdot7b^2=7b^2(2-5b)\\\\2.) 18x^3y^2 + 42x^2y^3=3x\cdot6x^2y^2+7y\cdot6x^2y^2=6x^2y^2(3x+7y)\\\\3.) 12x^2 + 16y^2=4(3x^2+4y^2)\\\\4.) 3x^2 - 9xy + 12y=3(x^2-3xy+4)\\\\5.) 45x^3 - 60xy^2=3x^2\cdot15x-4y^2\cdot15x=15x(3x^2-4y^2)\\\\6.)36x^2+18x^3-6x+12x^4=6x(2x^3+3x^2+6x-1)$

$Ans.\ GFC:\ 1)\ 7b^2,\ 2)\ 6x^2y^2,\ 3)\ 4,\ 4)\ 3,\ 5)\ 15x,\ 6)\ 6x$

natima [27]3 years ago

6 0

$1.\\14b^2-35b^3\\GCF(14;35)=7\\GCF(b^2;b^3)=b^2\\14b^2-35b^3=7b^2(2-5b)\\\\2.\\18x^3y^2+42x^2y^3\\GCF(18;42)=6\\GCF(x^3y^2;x^2y^3)=x^2y^2\\18x^3y^2+42x^2y^3=6x^2y^2(3x+7y)$

$3.\\12x^2+16y^2\\GCF(12;16)=4\\GCF(x^2;y^2)=1\\12x^2+16y^2=4(3x^2+4y^2)\\\\4.\\3x^2-9xy+12y\\GCF(3;9;12)=3\\GCF(x^2;xy;y)=1\\3x^2-9xy+12y=3(x^2-3xy+4y)$

$5.\\45x^3-60xy^2\\GCF(45;60)=15\\GCF(x^3;xy^2)=x\\45x^3-60xy^2=15x(3x^2-4y^2)\\\\6.\\36x^2+18x^3-6x+12x^4\\GCF(36;18;6;12)=6\\GCF(x^2;x^3;x;x^4)=x\\36x^2+18x^3-6x+12x^4=6x(6x+3x^2-1+2x^3)$

$GCF:\\1)\ b^2;\ 2)\ 6x^2y^2;\ 3)\ 4;\ 4);\ 3;\ 5);\ 15x;\ 6)\ 6x$

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A normal population has a mean of 19 and a standard deviation of 5.

dangina [55]

Answer:

a) $Z = 1.2$

b) 38.49% of the population is between 19 and 25.

c) 34.46% of the population is less than 17.

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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. If we need to find the probability that the measure is larger than X, it is 1 subtracted by this pvalue.

For this problem, we have that

A normal population has a mean of 19 and a standard deviation of 5, so $\mu = 19, \sigma = 5$ .