Marcus was training for a 2,000 meter race. He had to alternate between running and walking.

The 8th grade is selling tickets to the 8th grade formal.The cost per ticket is $15.The decorations and food cost $600,and each

MArishka [77]

Answer:

Step-by-step explanation:

Each ticket is $15. The number of tickets is what we are trying to solve for. The class spends a certain amount of money to prepare for the formal. They hope that the money they make in ticket sales is MORE than what they spend. The expression that represents the number of tickets at $15 each is 15x, where x is the number of tickets. They hope that the sales are greater than what they spend, so what we have so far is

15x >

Greater than what, though? What do they spend? They spend 600 for the food, so

15x > 600...

but they also have to print a certain, unknown number of tickets at .50 each. The expression that represents the printing of each ticket is .5x (we can drop the 0; it doesn't change the answer or make it wrong if we drop it off). So the cost for this affair is the food + the printing.

15x > 600 + .5x

Solve this inequality for x. Begin by subtracting .5 from both sides to get

14.5x > 600 so

x > 41.3

Because we are not selling (or printing) .3 of a ticket, it's safe to say (and also correct!) that they need to sell (and print) 41 tickets. If they sell 41 tickets, the profit is found by

15(41) > 600 + .5(41)

615 > 600

This means that at 41 tickets, they make a profit. At 40 tickets, the inequality looks like this:

15(40) > 600 + .5(40) and

600 > 620. This is not true, so 40 tickets isn't enough.

8 0

3 years ago

A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of

finlep [7]

Answer:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

$p_v =2*P(t_{(7)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=0.475$

The sample deviation calculated $s=0.183$

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 0.5$

Alternative hypothesis: $\mu \neq 0.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=8-1=7$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(7)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

4 0

2 years ago

Read 2 more answers

Sally has a discount card that reduces the price of her grocery bill in a certain grocery store by 5%. If c represents the cost

amm1812

The original price is c. Then the grocery bill will be: c - 0.05c = 0.95 c. You can also say since the beggining that 5% disscount implies 0.95 factor, which leads to c * 0.95 = 0.95c. Sorry if it’s wrong

3 0

2 years ago

12a^2*b^4-36a^2*b+44abc

spayn [35]

Let's simplify step-by-step.

$12a^2*b^4-36a^2*b+44abc=$

There are no like terms.

Answer:

$12a^2*b^4-36a^2*b+44abc=$

5 0

2 years ago

Help I need to turn this in by 12:30

11Alexandr11 [23.1K]

B- -3

If x is two according to the coordinate points, that means the equation is now -8+3y=-17

Add 8 to each side- 3y=-9

And divide -9 by 3

This gets you y=-3

5 0

2 years ago