What is the answer? 102,39,or 78

Which shows a correct order to solve this story problem? Kent and Curtis went to the state fair. They had to pay a total of $7.1

KiRa [710]

The problem is asking how much each person will need to pay. Simplifying the problem into an equation with variables (an algorithm) will greatly help you solve it:

S = Sales Tax = $ 7.18 per any purchase
A = Admission Ticket = $ 22.50 entry price for one person (no tax applied)
F = Food = $ 35.50 purchases for two people

We know the cost for one person was: (22.50) + [(35.50/2) + 7.18] =
$ 47.43 per person. Now we can check each method and see which one is the correct algorithm:

Method A)
[2A + (F + 2S)] / 2 = [ (2)(22.50) + [35.50 + (2)(7.18)] ]/ 2 = $47.43
Method A is the correct answer

Method B)
[(2A + (1/2)F + 2S) /2 = [(2)(22.50) + 35.50(1/2) + (2)7.18] / 2 = $38.55
Wrong answer. This method is incorrect because the tax for both tickets bought are not being used in the equation.

Method C)
[(A + F) / 2 ]+ S = [(22.50 + 35.50) / 2 ] + 7.18 = $35.93
Wrong answer. Incorrect Method. The food cost is being reduced to the cost of one person but admission price is set for two people.

6 0

3 years ago

Order 51.835, 51.836, 51.736, and 51.837 in decreasing order.

Tju [1.3M]

Answer:

D.)51.837, B.)51.836, A.)51.835, C.)51.736

Step-by-step explanation:

its decreasing in this order

3 0

3 years ago

Read 2 more answers

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago

These triangles are similar. find Y and X

CaHeK987 [17]

It these triangles are similar, then the sides of the triangles are in proportion:

$\dfrac{y}{10}=\dfrac{y+4}{15}$ cross multiply