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11111nata11111 [884]
3 years ago
14

What is the hydrogen ion concentration of orange juice that has a pH of 1?

Mathematics
1 answer:
saw5 [17]3 years ago
5 0

Answer:

0.1 mole per liter

Step-by-step explanation:

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The solution is (6, -8) (solved by graphing with Desmos.com/calculator)
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Which equivalent expression would you set up to verify the associative property of addition for (3x + 4) + ((5x2 – 1) + (2x + 6)
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(3x+4)+(5x2-1)+(2x+6)

remove unnecessary parenthesis

3x+4+(5x2-1)+(2x+6)

3x+4+(10-1)+2x+6

subtract the numbers

3x+4+9+2x+6

collect the like terms

5x+4+4+9+6

5x+19

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On a certain hot​ summer's day, 769 people used the public swimming pool. The daily prices are $1.25 for children and $2.50 for
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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.

3 0
3 years ago
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