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3 years ago

If the mean of symmetrical distribution is 28, which of these values could be the median of the distribution.

Mathematics

1 answer:

lina2011 [118]3 years ago

6 0

If a distribution is symmetric, its mean is the same as its median, so the median would also be 28.

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I know I need to use the change of base

Zinaida [17]

The change of base formula looks like this:
log x
(log to the base b of ) x = ----------- (you could also use ln instead of log)
log b

log 40 1.602
Then (log to the base 7 of) 40 = ------------ = ------------- = 1.896 (answer)
log 7 0.8451

check (on calculator): 7^1.896 = 40.023 (close enough to 40 to verify
that this method "worked")

7 0

3 years ago

TIMED PLEASE HURRY HELP WITH 2 EASY QUESTIONS

Alex_Xolod [135]

<h2> Question # 1</h2><h2>Which statements are true?</h2><h2 /><h3><u>Analyzing and solving the first statement:</u></h3>

$4g^2-g=g^2\left(4-g\right)$

Solving the expression

$4g^2-g$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^2=gg$

So,

$4gg-g$

$\mathrm{Factor\:out\:common\:term\:}g$

$g\left(4g-1\right)$

So,

$4g^2-g:\quad g\left(4g-1\right)$

Therefore, the statement $4g^2-g=g^2\left(4-g\right)$ is NOT CORRECT.

<h3><u>Analyzing and solving the second statement:</u></h3>

$35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Solving the expression

$35g^5-25g^2$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^5=g^3g^2$

So,

$35g^3g^2-25g^2$

$\mathrm{Rewrite\:}25\mathrm{\:as\:}5\cdot \:5$

$\mathrm{Rewrite\:}35\mathrm{\:as\:}5\cdot \:7$

$5\cdot \:7g^3g^2-5\cdot \:5g^2$

$\mathrm{Factor\:out\:common\:term\:}5g^2$

$5g^2\left(7g^3-5\right)$

So,

$35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Therefore, the statement $35g^5-25g^2=\:5g^2\left(7g^3-5\right)$ is CORRECT.

<h3><u>Analyzing and solving the third statement:</u></h3>

$24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$

<h3 />

Solving the expression

<h3> $24g^4+18g^2$ </h3><h3> $24g^2g^2+18g^2$ </h3><h3> $\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$ </h3><h3> $\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$ </h3><h3> $6\cdot \:4g^2g^2+6\cdot \:3g^2$ </h3><h3> $\mathrm{Factor\:out\:common\:term\:}6g^2$ </h3><h3> $6g^2\left(4g^2+3\right)$ </h3>

So,

<h3> $24g^4+18g^2=6g^2\left(4g^2+3\right)$ </h3>

Therefore, the statement $24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$ is CORRECT.

<h3><u>Analyzing and solving the fourth statement:</u></h3>

$9g^3+12=\:3\left(3g^3+4\right)$

Solving the expression

$9g^3+12$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}3\cdot \:4$

$\mathrm{Rewrite\:}9\mathrm{\:as\:}3\cdot \:3$

$3\cdot \:3g^3+3\cdot \:4$

$\mathrm{Factor\:out\:common\:term\:}3$

$3\left(3g^3+4\right)$

So,

$9g^3+12=\:3\left(3g^3+4\right)$

Therefore, the statement $9g^3+12=\:3\left(3g^3+4\right)$ is CORRECT.

<h2> Question # 2</h2><h2>Which expressions are completely factored?</h2>

<u>Solving first expression</u>

Considering the expression

$30a^6-24a^2$

$30a^6-24a^2$

$30a^4a^2-24a^2$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$\mathrm{Rewrite\:}30\mathrm{\:as\:}6\cdot \:5$

$6\cdot \:5a^4a^2-6\cdot \:4a^2$

$\mathrm{Factor\:out\:common\:term\:}3a^2$

$3a^2\left(10a^4-8\right)$

Thus, the expression $30a^6-24a^2=3a^2\left(10a^4-8\right)\:$ is completely factored.

<u>Solving second expression</u>

Considering the expression

$12a^3-8a$

$12a^3-8a$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^3=a^2a$

So,

$12a^2a-8a$

$\mathrm{Rewrite\:}8\mathrm{\:as\:}4\cdot \:2$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}4\cdot \:3$

$4\cdot \:3a^2a-4\cdot \:2a$

$\mathrm{Factor\:out\:common\:term\:}4$

$4\left(3a^3-2a\right)$

Thus, the expression $12a^3-8a=\:4\left(3a^3-2a\right)$ is completely factored.

<u>Solving third expression</u>

$16a^5-20a^3\:\:\:\:\:\:\:\:\:\:\:\:$

$16a^5-20a^3$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^5=a^2a^3$

So,

$16a^2a^3-20a^3$

$\mathrm{Rewrite\:}20\mathrm{\:as\:}4\cdot \:5$

$\mathrm{Rewrite\:}16\mathrm{\:as\:}4\cdot \:4$

$4\cdot \:4a^2a^3-4\cdot \:5a^3$

$\mathrm{Factor\:out\:common\:term\:}4a^3$

$4a^3\left(4a^2-5\right)$

Thus, the expression $16a^5-20a^3\:=4a^3\left(4a^2-5\right)$ is completely factored.

<u>Solving fourth expression</u>

$24a^4+18$

$24a^4+18$

$\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$6\cdot \:4a^4+6\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}6$

$6\left(4a^4+3\right)$

Thus, the expression $24a^4+18=6\left(4a^4+3\right)$ is completely factored.

Keywords: expression, factoring

Learn more about expression factoring from brainly.com/question/14051207

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8 0

3 years ago

What equation results from completing the square and then factoring?<br> x2 + 10x = 15

Stells [14]

12x=15
Divide by 12
X=5/4

6 0

3 years ago

Read 2 more answers

An executive invests $26,000, some at 8% and some at 6% annual interest. If he receives an annual return of $1,860, how much is

S_A_V [24]

Let x = amount invested at 8%
26000 - x = amount invested at 6%

0.08x + 0.06(26000 - x) = 1860

solving the equation you get x = 15000
and subtract that from 26000 you get 11000
8% investment = 15000
6% investment = 11000

5 0

3 years ago

Which polynomial is a product of (3x+2) and (x-7)

pogonyaev

(3x+2) (x-7)

FOIL

first 3x*x = 3x^2

outer 3x*-7 = -21x

inner = 2*x = 2x

last 2*-7 = -14

add them together = 3x^2 -21x+2x -14

combine like terms 3x^2 -19x -14

5 0

3 years ago