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2 years ago

Plzzzz answer this, its urgent!!!!!

1 answer:

7 0

1. multiply by 1.046
2. 500* 1.046=523
3. 6*(6/10)=(36/10)=3.6
4. 0.5
5. 34.5714+17.43=52
6. 2
7. x=1
8. 5n-2
9. -9
10. 12.6

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Step-by-step explanation:

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2 years ago

How many units up was the graph of f(x) =∛x shifted to form the translation?

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3 years ago

Read 2 more answers

Tacoma's population in 2000 was about 200 thousand, and has been growing by about 8% each year. If this continues, what will Tac

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Answer: The population will be 408,000 people.

Step-by-step explanation:

So in 2000 there were 200,000 people and it started to grow 8% every year so up to 2013.

so find 8% of 200,000 and then multiply it by the the number of years.

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2 years ago

Use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come f

AlladinOne [14]

Answer:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

Step-by-step explanation:

Data given and notation

n=75 represent the random sample taken

$\hat p=0.64$ estimated proportion of interest

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true proportion is higher than 0.5:

Null hypothesis: $p =0.5$

Alternative hypothesis: $p > 0.5$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

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