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grin007 [14]
3 years ago
12

1) What is the scale factor of dilation when the blue shape is a dilation of the black shape?

Mathematics
1 answer:
Anon25 [30]3 years ago
4 0
B and A is the correct answers
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If M={a,b,c,d} and N={c,b,e,f},find the value of M-N​
kumpel [21]

Answer:

M-N= {a,d}

Step-by-step explanation:

subtract the common elements

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I have no idea what to do with this someone help
joja [24]
The given distance of 5.20 would be A.

Replace A with 5.20 to solve for T.

T = 5.20^3/2
T = 11.9 years.
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Acute angles Angles that measure 60°
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Acute angles are angles that measure below 90°, meaning below a right angle.
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An ulcer patient has been told to avoid acidic foods. If he drinks coffee, with a pH of 5.0, it bothers him, but he can tolerate
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The equation should look like :
[H+] = (1-a)(x) + (a)(y)
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3 0
3 years ago
A recent broadcast of a television show had a 10 ​share, meaning that among 6000 monitored households with TV sets in​ use, 10​%
Kisachek [45]

Answer:

Null and alternative hypothesis

H_0: \pi \geq0.25\\\\H_1: \pi

Test statistic z=-26.82

P-value P=0

The null hypothesis is rejected.

It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

Step-by-step explanation:

We have to perform a hypothesis test of a proportion.

The claim is that less than 25% were tuned into the program, so we will state this null and alternative hypothesis:

H_0: \pi \geq0.25\\\\H_1: \pi

The signifiance level is 0.01.

The sample has a proportion p=0.1 and sample size of n=6000.

The standard deviation of the proportion, needed to calculate the test statistic, is:

\sigma_p=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.25(1-25)}{6000} } =0.0056

The test statistic is calculated as:

z=\frac{p-\pi+0.5/N}{\sigma_p} =\frac{0.1-0.25+0.5/6000}{0.0056}=\frac{-0.1499}{0.0056}  =-26.82

As this is a one-tailed test, the P-value is P(z<-26.82)=0. The P-value is smaller than the significance level (0.01), so the effect is significant.

Since the effect is significant, the null hypothesis is rejected. It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

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2 years ago
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