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Nesterboy [21]
3 years ago
15

32.25 written as a mixed number

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
7 0
Answer: 32 1/4

That is the whole number——>32 1/4 <—— that is the fraction
xenn [34]3 years ago
7 0
32 1/4 is the correct answer
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Answer: 3 cakes were at the party

Step-by-step explanation:

Multiply 1/8*24 to find the exact number of cakes that were at the party

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admissions to a carnival is $5 and each game at the carnival costs 0.85 you have $15 to spend on admissions and games what is th
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You have $15 and admission is $5, so you know have $10 before doing any games. If you wanted to play 12 games, you would need $10.20, which you do not have. But, if you wanted to play 11 games, then you would need $9.35, which is just under $10.

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8 0
4 years ago
Simplify (3x – 5) + (5x + 1).
Andrej [43]

Answer:

(-2x - 6)

Step-by-step explanation:

The given expression is ( 3x - 5 ) - ( 5x + 1 )

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( 3x - 5 ) - 5x - 1

3x - 5 - 5x - 1

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3x - 5x - 5 -1

-2x - 6

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3 years ago
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Find the square root of the following decimal numbers.​
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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.

3 0
3 years ago
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