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3 years ago

5.A Nathan decided he wanted to go to a theme park. The theme

Mathematics

1 answer:

rjkz [21]3 years ago

3 0

Answer:

5a. $1.50R + $7

5b. 14 rides

Step-by-step explanation:

5b. $28 - $7= $21/1.50= 14

He can go on 14 rides

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I’m totally confused can someone please help me!!!

german

The second answer is right

3 0

3 years ago

Read 2 more answers

Last year 64 new houses were built in a neighborhood so far this year 7/8 of them have been sold how many new houses have been s

laiz [17]

Answer:

Step-by-step explanation:

for this problem you would first figure out what 1/8 of 64 is. 1/8 of 64 is 8.so we do 7*8 to get 56 so 56 of the 64 houses have been sold

6 0

2 years ago

Given the domain (-2,2,4) what is the range for the relation 3x-y=3

Bumek [7]

Answer:

Given the domain, the range for 3x-y = 3 is {-9, 6, 9}

Step-by-step explanation:

First you have to put the relation in terms of y ⇒ 3x - y = 3⇒ 3x -3 = y

⇒ y = 3x - 3.

Then you replace the values indicated by the domain to find their "y" values (the ones that constitute the range).

f(-2) = -9

f(2) = 6

f(4) = 9.

Finally, the range for the given domain is {-9, 6, 9}

3 0

3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.