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devlian [24]
2 years ago
6

Maya shaved her head and then began letting her hair grow.

Mathematics
2 answers:
Brums [2.3K]2 years ago
6 0

Answer:

Maya grows hair

Step-by-step explanation:

Alexxandr [17]2 years ago
3 0

Answer:Maya will not be bald anymore.

Step-by-step explanation:

Her hair is growing. sCiEnCe

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You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 nigh
lorasvet [3.4K]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean GPA of night students and μ2 be the mean GPA of day students.

The random variable is μ1 - μ2 = difference in the mean GPA of night students and the mean GPA of day students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 2.35

x2 = 2.58

s1 = 0.46

s2 = 0.47

n1 = 30

n2 = 25

t = (2.35 - 2.58)/√(0.46²/30 + 0.47²/25)

t = - 1.82

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.46²/30 + 0.47²/25]²/[(1/30 - 1)(0.46²/30)² + (1/25 - 1)(0.47²/25)²] = 0.00025247091/0.00000496862

df = 51

We would determine the probability value from the t test calculator. It becomes

p value = 0.075

Alpha = 5% = 0.05

Since alpha, 0.05 < than the p value, 0.075, then we would fail to reject the null hypothesis.

4 0
2 years ago
An angle t is drawn from the center of the unit circle. Find a formula in terms of t for the straight line distance d between th
Doss [256]

Answer:

d = \sqrt{2}\cdot \sqrt {1-\cos t - \sin t}

Step-by-step explanation:

Let suppose that one of the radii meets the circle at the point (1,0). The straight line distance formula is:

d = \sqrt{(\cos t - 1)^{2}+(\sin t - 1)^{2}}

d = \sqrt{(\cos^{2}t - 2\cdot \cos t + 1)+(\sin^{2}t - 2\cdot \sin t + 1)}

d = \sqrt{2-2\cdot (\cos t + \sin t )}

d = \sqrt{2}\cdot \sqrt {1-\cos t - \sin t}

5 0
2 years ago
Round 22360 to 1 significant figure
monitta
This link to a video will help

https://youtu.be/k3RwfEvqrB8
7 0
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If f Superscript negative 1 Baseline (x) = negative one-fifth x, what is f Superscript negative 1 Baseline (x) = one-fifth x?
Artyom0805 [142]

Answer:

  a different inverse function

Step-by-step explanation:

If f^{-1}(x)=-\frac{1}{5}x then f^{-1}(x)=\frac{1}{5}x is a different inverse function.

__

The second (inverse) function is the first reflected over the y-axis.

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There are 180 pages in the book Diana is reading. Diana has read 90 pages. What percent of the book has she read?
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The answer is fifty percent.
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2 years ago
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