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Gemiola [76]
4 years ago
9

PLEASE HELP U WILL GET 15 POINTS A quadrilateral labeled figure A. The sides are labeled 4 centimeters, 2 centimeters, 3 centime

ters, and 5 centimeters, clockwise about the figure. What figure is a dilation of Figure A by a factor of 2? Note that the images are not necessarily drawn to scale. A quadrilateral. The sides are labeled 2 centimeters, 1 centimeter, 1 and a half centimeters, and 2 and a half centimeters, A quadrilateral. The sides are labeled 6 centimeters, 4 centimeters, 5 centimeters, and 7 centimeters, clockwise about the figure. A quadrilateral. The sides are labeled 2 centimeters, one-half centimeter, 1 centimeter, and 3 centimeters, clockwise about the figure. A quadrilateral. The sides are labeled 8 centimeters, 4 centimeters, 6 centimeters, and 10 centimeters, clockwise about the figure.
Mathematics
1 answer:
Eva8 [605]4 years ago
3 0
The sides are 2, 1, 1.5, and 2.5 clockwise
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3 years ago
Find a + b, if b^a = 64 and a^b = 81
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\bf \begin{array}{llll}
b^a=64\qquad \boxed{64=4^3}\qquad b^a=4^3
\\\\\\
a^b=81\qquad \boxed{81=3^4}\qquad a^b=3^4
\end{array} \qquad 
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a=3\\
b=4
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3 years ago
What percent of 1/2 is 2/5
eimsori [14]
1/2= 50%
2/5=40% is the answer
4 0
3 years ago
If c = the number of cars in a parking lot, which algebraic expression
stira [4]

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9c

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8 0
4 years ago
Read 2 more answers
Find the P-value for a test of the claim that more than 60% of the people following a particular diet will experience increased
lina2011 [118]

Answer:

The p-value of the test is 0.0207.

Step-by-step explanation:

Test of the claim that more than 60% of the people following a particular diet will experience increased energy (H1: p > 0.6).

The null hypothesis is:

H_0: p = 0.6

The alternate hypothesis is:

H_1: p > 0.6

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.6 is tested at the null hypothesis:

This means that \mu = 0.6, \sigma = \sqrt{0.6*0.4}

Of 100 randomly selected subjects who followed the diet, 70 noticed an increase in their energy level.

This means that n = 100, X = \frac{70}{100} = 0.7

Test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.7 - 0.6}{\frac{\\sqrt{0.6*0.4}}{\sqrt{100}}}

z = 2.04

P-value of the test:

The p-value of the test is the probability of finding a sample proportion of at least 0.7, which is 1 subtracted by the p-value of z = 2.04.

Looking at the z-table, z = 2.04 has a p-value of 0.9793

1 - 0.9793 = 0.0207

The p-value of the test is 0.0207.

4 0
3 years ago
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