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cluponka [151]
3 years ago
15

Acellus

Mathematics
1 answer:
Over [174]3 years ago
7 0
2/12 = .1666666 then do .1666666 times 15. That gives you 2.5 cups
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What are the sine, cosine, and tangent of Θ = 3 pi over 4 radians?
Daniel [21]
Our angle teta is:
teta = 3pi/4

since that is larger than pi/2 but less than pi that means that our angle lies in II quadrant (x negative y positive)

sin(3pi/4) = √2/2
cosine and tangent of that angle must be negative because of position of the angle.

cos(3pi/4) = -√2/2
tan(3pi/4) = -1
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Expand

12x - 48 + 4x + 28

Collect like terms

(12x + 4x) + (-48 + 28)

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A sample of 38 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.
kaheart [24]

Answer:

a. two-tailed test

b. reject H0 if Z>1.96 or Z<-1.96

c. Z = 1.73

d. we have failed to reject the null hypothesis

e. P-value = 0.0418

Step-by-step explanation:

We will use a Z test to resolve this, an it will be a two-tailed test because the hypothesis statements are not indicating a specific direction for the significant difference (H0 : μ1 = μ2 ; H1 : μ1 ≠ μ2), this also means that the significanclevel will be divided between the both tails (2.5% en each tail for the rejection regions). See attached drawing for reference.

We need to find our critical value:

Zα/2 = Z(0.05/2) = 0.025

If we look for 0.025 in a Z table we will find that the critical value is 1.96 to the right, and by symmetry -1.96 to the left. So our decision rule will be to reject H0 if Z>1.96 or Z<-1.96

The Z test will be done using the next equation:

Z = (x⁻1 - x⁻2) - (μ1 - μ2) / √( σ²1/n1 + σ²2/n2

Because we are testing the null hypothesis we know that μ1 - μ2 must be zero if they are supposed to be equal (H0 : μ1 = μ2), so we calculate as follows:

Z = (100.5 - 98.8) - (0) / √( (3.4)²/38 + (5.8)²/51 = 1.7/0.9817 = 1.73

Z<1.96, therefore we have failed to reject the null hypothesis

The P-value for this test would be represented by the probability of Z being greater than 1.73, so we can look for it in any Z table, finding that its value is 0.0418, and because P-value>0.025 we again confirm that we don't have evidence statistically significant to reject the null hypothesis

4 0
3 years ago
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