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3 years ago

7b³-9b=10 whats the answer for this

Mathematics

1 answer:

Svet_ta [14]3 years ago

6 0

Answer:

B just took it :)

Step-by-step explanation:

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Factor 4x 2 + 12x + 5 . (2x + 5)(2x + 1)

Juliette [100K]

Factor
in form
ax²+bx+c form
use the ac method
1. mulitply a and c
2. factor the result such that the 2 factors add to b
3. split b into those 2 factors
4. group the terms
5. undistribute common factors
6. undistribute again

first multiply ac
4 times 5=20
now
what 2 numbers multiply to 20 and add to 12
2 and 10
4x²+2x+10x+5
group
(4x²+2x)+(10x+5)
undistribute
2x(2x+1)+5(2x+1)
undistribute the (2x+1)
(2x+1)(2x+5) is factored form whih you have there, nice

8 0

3 years ago

Find d<br><br> a * a * a = 216<br> a * b * c = 96<br> b * c = 52<br> a + b + c = d

Slav-nsk [51]

$a\cdot a\cdot a=216\\\\a^3=216\to a=\sqrt[3]{216}\\\\\boxed{a=6}\\\\\text{Substitute}\ b\cdot c=52\ \text{to the second expression}\ a\cdot b\cdot c=96:\\\\abc96\ \wedge\ bc=52\to a(52)=96\qquad\text{divide both sides by 52}\\\\a=\dfrac{96}{52}\to a=\dfrac{24}{13}\neq6$

a = 6 and a = 24/13 FALSE!!!

<h3>Answer: NO SOLUTION.</h3>

3 0

4 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

Simplify the following: (-83)2

Anika [276]

The answer is -166, your welcome

8 0

3 years ago

What is the exsponet of 64/81 = 8/9

Mumz [18]

(8/9)^2 is 64/81. Exponent is 2. Hope this help

4 0

3 years ago