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3 years ago

Factor completely and then place the factors in the proper location on the grid.

2 answers:

8 0

1. 8c^2-26c+15= (4c-3) (2c-5). Break the expression into groups: =(8c^2-6c)+(-20c+15). Factor out 8c^2-6c: 2c(4c-3). Factor out -5 from -20c+ 15: -5(4c-3). Lastly factor out common term (4c-3) and thats how you'll get your answer (4c-3) (2c-5).

2. common factors for 270 and 360 is 90.To find this write the factors of each and find the largest one.270: 1, 270, 2, 135, 3, 90, 5, 54, 6, 45, 9, 30, 10, 27, 15, 18360: 1, 360, 2, 180, 3, 120, 4, 90, 5, 72, 6, 60, 8, 45, 9, 40, 10, 36, 12, 30, 15, 24, 18, 20
3. The factors for 8 a3b2 and 12 ab4 is 4. because 8: 1, 2, 4, 812: 1, 2, 3, 4, 6, 12.

4. 81a^2+36a+4= (9a+2)^2. Break down the expression into groups: (81a^2+18a)+(18a+4). Factor out 9a from 81a^2 +18a: 9a(9a+2). Factor out 2 from 18a+4: 2(9a+2). so the groups you got are now 9a(9a+2)+2(9a+2). Lastley factor out common term (9a+2) to get (9a+2) (9a+2). Finally you get the answer (9a+2)^2.

5. mn-15+3m-5n= (n+3)(m-5). factor out m from nm+3m: m(n+3). Factor out -5 from -5n-15: -5(n+3). And thats how you get the number (n+3)(m-5)

Hope this helped :) Have a great day

notsponge [240]3 years ago

3 0

Answer:

1. $(2c-5)(4c-3)$

2. $2\times 3^2\times 5$

3. $4ab^2$

4. $(9a+2)(9a+2)$

4. $(m-5)(n+3)$

Step-by-step explanation:

1.We are given that

$8c^2-26c+15$

We have to find the factor

Using factorization method

$8c^2-20c-6c+15$

$4c(2c-5)-3(2c-5)$

$(2c-5)(4c-3)$

2.We are given that 270 and 360

We have to find the HCF of 270 and 360

$270=2\times 3^3\times 5$

$360=2^3\times3^2\times 5$

Hence, HCF of 270 and 360 = $2\times 3^2\times 5$

3.We are given that $8a^3b^2$ and $12ab^4$

We have to find HCF

$8a^3b^2=2^3\times a\times a\times a\times b\times b$

$12 ab^4=2^2\times 3\times a\times b\times b\times b\times b$

Common factors are 4,a, $b^2$

Hence,HCF $8a^3b^2$ and $12 ab^4 =4ab^2$

4. $81a^2+36a+4$

We have to find the factors

$81a^2+18a+18a+4$

$9a(9a+2)+2(9a+2)$

$(9a+2)(9a+2)$

5. $mn-15+3m-5n$

We have to find the factors

$mn+3m-5n-15$

$m(n+3)-5(n+3)$

$(m-5)(n+3)$

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2 years ago

Read 2 more answers

Claim: The standard deviation of pulse rates of adult males is less than 11 bpm. For a random sample of 132 adult males, the pu

seraphim [82]

Answer:

$\chi^2 =\frac{132-1}{121} 114.49 =123.952$

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=132$ represent the sample size

$\alpha$ represent the confidence level

$s^2 =10.7^2 =114.49$ represent the sample variance obtained

$\sigma^2_0 =11^2=121$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is violated, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 121$

Alternative hypothesis: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{132-1}{121} 114.49 =123.952$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 131. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2$