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3 years ago

BRAINLIEST!!!!!!BRAINLIEST!!!!BRAINLIEST!!!!BRAINLIEST!!!BRAINLIEST!!!!!BRAINLIEST!!!!BRAINLIEST!!!!BRAINLIEST !!!!!!!

Mathematics

2 answers:

faust18 [17]3 years ago

8 0

Answer:

answer C

Step-by-step explanation:

$3x^2-4x-8 + (6x^2+2x-3) = 3x^2-4x-8+6x^2+2x-3=9x^2-2x-11$

so the correct answer is the C

hope this helps

Airida [17]3 years ago

8 0

Answer:

Step-by-step explanation:

= 3x²-4x-8+6x²+2x-3

= 3x²+6x²-4x+2x-8-3

= 9x²-2x-11

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3 years ago

Researchers studying starfish collected two independent random samples of 40 starfish. One sample came from an ocean area in the

krek1111 [17]

Answer:

The test statistic for the appropriate test is $\frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}$ .

Step-by-step explanation:

The experiment conducted here is to determine whether there is a difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. north and south.

The hypothesis to test this can be defined as follows:

H₀: There is no difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. p₁ = p₂.

Hₐ: There is a difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. p₁ ≠ p₂.

The two-proportion z-test would be used to perform the test.

A sample of n = 40 starfishes are selected from both the ocean areas.

It provided that of the 40 starfish from the north, 6 were found to be over 8 inches in length and of the 40 starfish from the south, 11 were found to be over 8 inches in length.

Compute the sample proportion of starfish from north that were over 8 inches in length as follows:

$\hat p_{n}=\frac{6}{40}=0.15$

Compute the sample proportion of starfish from south that were over 8 inches in length as follows:

$\hat p_{s}=\frac{11}{40}=0.275$

The test statistic is:

$z=\frac{\hat p_{n}-\hat p_{s}}{\sqrt{P(1-P)[\frac{1}{n_{n}}+\frac{1}{n_{s}}]}}$

Compute the combined proportion P as follows:

$P=\frac{X_{n}+X_{s}}{n_{n}+n_{s}}=\frac{6+11}{40+40}=0.2125$

Compute the test statistic value as follows:

$z=\frac{\hat p_{n}-\hat p_{s}}{\sqrt{P(1-P)[\frac{1}{n_{n}}+\frac{1}{n_{s}}]}}$

$=\frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}$

Thus, the test statistic for the appropriate test is $\frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}$ .

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3 years ago