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3 years ago

11

Miranda was babysittingfor a friends parents saturday morning and was paid $12.50 per hour. Later that evening, she was babysitt

ing for another couple that paid her $18.75 per hour for 5 hours. What was mirandas average rate of pay per hour for the day?

1 answer:

Shkiper50 [21]3 years ago

3 0

Answer:

10-20 dollars per hour

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Find the 95% confidence interval for estimating the population mean μ

AVprozaik [17]

We first need to determine whether we are dealing with means or proportions in this problem. Since we are given the sample and population mean, we know that we are dealing with means.

Since we have one sample mean, this means we are creating a confidence interval for one sample (1 Samp T Int).

Normally we would check for conditions, but since this is not formulated as a "real-world scenario" type problem, it is hard to check for randomness and independence. Therefore, I will be excluding conditions from this answer.

<h3>Confidence Interval Formula</h3>

The formula for constructing a <u>confidence interval for means</u> is as follows:

$\displaystyle \overline{x} \pm t^*\big{(}\frac{\sigma}{\sqrt{n} } \big{)}$

We are given these variables:

$\overline{x}=50$
$n=60$
$\sigma=10$

Plug these values into the formula for the confidence interval:

$\displaystyle 50\pm t^* \big{(}\frac{10}{\sqrt{60} } \big{)}$

<h3>Finding the Critical Value (t*)</h3>

In order to find t*, we can use this formula:

$\displaystyle \frac{1-C}{2}=A$

Calculating the z-score associated with "A" will give us t*.

So, let's plug in the confidence interval 95% (.95) into the formula:

$\displaystyle \frac{1-.95}{2}=.025$

Use your calculator or a t-table to find the z-score associated with this area under the curve.. you should get:

$t^*=1.96$

<h3>Constructing Confidence Interval</h3>

Now, let's finish the confidence interval we created:

$\displaystyle 50\pm 1.96 \big{(}\frac{10}{\sqrt{60} } \big{)}$

We can calculate the confidence interval, using this formula, to be:

$\boxed{(47.4697, \ 52.5303)}$

<h3>Interpreting the Confidence Interval</h3>

We are 95% confident that the true population mean μ lies between <u>47.4697 and 52.5303</u>.

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C and D

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3 years ago

Use special right triangle ratios to find the length of the hypotenuse.

Nady [450]

Answer:

Step-by-step explanation:

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3 years ago

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A ran

mart [117]

Answer:

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

$X=48$ number of vehicles classified as sports utility.

$n=500$ random sample taken

$\hat p=\frac{48}{500}=0.096$ estimated proportion of vehicles classified as sports utility vehicles.

$p$ true population proportion of vehicles classified as sports utility vehicles.

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".

The population proportion have the following distribution

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

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

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 $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070$

$0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121$

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

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3 years ago

Which expression has the same value as the one below? <br> 29+(-37)

Dmitry [639]

Answer:

B

Step-by-step explanation:

29 + (-37) = 29 -37

Leave a like and mark brainliest if this helped

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3 years ago

Read 2 more answers