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3 years ago

What is the slope for y = 4 + 3x? * Your answer

Mathematics

1 answer:

Leto [7]3 years ago

5 0

Answer:

Step-by-step explanation:

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Multiply and simplify

steposvetlana [31]

That would be the cube root of (x+5)^11.

If desired, this could be reduced to the cube root of (x+5)^9*(x+5)^2, which would be

(x+5)^3*(x+5)^(2/3)

7 0

4 years ago

Solve the system of equations x + y = 4 and -9x + 7y = -84 by combining the equations.

Yanka [14]

Answer:

x = 7

y = -3

Step-by-step explanation:

x + y = 4 (Multiply this by 9)

-9x + 7y = -84

9x + 9y = 36

-9x + 7y = -84

16y = -48

y = -3

x + y = 4

x - 3 = 4

x = 7

6 0

3 years ago

Suppose that circles A and B have a central angle measuring 100°. Additionally, the measure of the sector for circle A is 10π m2

34kurt

12 m

10π40π = 62x2
x = 12

When circles have the same central angle measure, the ratio of measure of the sectors is the same as the ratio of the radii squared.

3 0

4 years ago

Read 2 more answers

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago

How can you write 1.121212....has a mix number

balu736 [363]

Look what I drew for you and it is your answer

8 0

4 years ago