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3 years ago

12

It cost $13 for admission to an amusement park, plus $1.50 for each ride. If you have a total of $33.50 to spend, what is the gr

eatest number of tides you can go on

2 answers:

svet-max [94.6K]3 years ago

8 0

13 rides is the greatest number of rides u can go on

Lana71 [14]3 years ago

8 0

See the image for the full explanation. Edit: I see you copied the total dollar cost incorrectly from the image to the question now that I look at it. This provides the perfect opportunity for you to try to get the correct answer using the methodology I described! (Hint: The answer is actually 15)

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What’s the slope of (5,3) and (10,8)

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The slope of this ordered pair is 1.

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For football tryouts at a local school, 12 coaches and 42 players will split into groups. Each group have the same number of coa

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<span>The biggest number of groups that can be formed is 6, because six is the greatest common factor of 12 and 42. There will be 2 coaches in each group because 6 x 2 = 12 and 7 players in each group because 7 x 6 = 42.</span>

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What is 6n-10=8 show me how to do this problem

Lady bird [3.3K]

Answer: n=3

Step-by-step explanation:

6n-10=8

6n=18

n=3

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3 years ago

Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

tigry1 [53]

Answer:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

$\bar X =237$ represent the sample mean

$\sigma = 47$ represent the population deviation

$n =30$ represent the sample size selected

$\mu_0 = 247$ represent the value that we want to test.

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

For the 90% confidence the value of the significance is given by $\alpha=1-0.9 = 0.1$ and $\alpha/2 = 0.05$ so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

$z_{\alpha/2}= 1.64$

And the margin of error would be:

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

The confidence interval for this case would be given by:

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

P-value

Since is a left tailed test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

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2 years ago

Jada walks up to a tank of water that can hold up to 15 gallons. When it is active, a drain empties from the tank at a constant

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Answer:

3 minutes 6 gallons of water is gone so therefor it would take 6 complete minutes but 3 more minutes

Step-by-step explanation:

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2 years ago