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kramer
3 years ago
6

Jason purchased a pack of game cards that was on sale for 13% off. The sales tax in his county is 8%. Let y represent the origin

al price of the cards. Write an expression that can be used to determine the final cost of the cards.
Mathematics
2 answers:
stealth61 [152]3 years ago
8 0

Answer:

0.08(0.87y)

Step-by-step explanation:

100% - 13% = 87% or 0.87

0.08 is the tax

0.08 (0.87y)

Plz give brainlest

DanielleElmas [232]3 years ago
5 0
I'm not completely sure, but I'm trying.
You have 13% off, so you multiply y by .87. Then, you multiply the tax and total (.87y×.08y) to get the product of .0696y. After that step, you add the total and the product(.87y×.0696) and get the sum of .939y which will be your final answer.

For the expression, you can write Jason can use .939y to find the final cost of the cards.
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Below is the data of ages at the local amusement park on a certain day.
nignag [31]

Answer:

39

Step-by-step explanation:

To find the intervals you will need to find the lowest and highest numbers, in this case, it would be 1 and 35.  A general rule would be 5-7 intervals, I will use 5.

Here are the intervals with the number of people that were in each:

1-7 (9)

8-14 (14)

15-21 (8)

22-28 (3)

29-35 (5)

14+9+8+3+5=39

6 0
3 years ago
Solve the equation. Show work for partial credit.
Natali5045456 [20]

Answer:

x = \frac{8}{5}

Step-by-step explanation:

Given

\frac{5}{x} + \frac{3}{x} = 5 ← combine terms on left side

\frac{8}{x} = 5 ( multiply both sides by x )

8 = 5x ( divide both sides by 5 )

\frac{8}{5} = x

8 0
3 years ago
A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research fir
AVprozaik [17]

Answer:

a) F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be \alpha/2 =0.025, the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: F>4.026 \cup F since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation  

n_1 = 10 represent the sampe size for the Miller's Stores

n_2 =10 represent the sample size for the Albert's stores

\bar X_1 =121.92 represent the sample mean for Miller's store

\bar X_2 =114.81 represent the sample mean for Albert's store

s_1 = 1.4 represent the sample deviation for the Miller's store

s_2 = 1.84 represent the sample deviation for the Albert's stores

s^2_2 = 12.25 represent the sample variance for the utility stocks

\alpha=0.05 represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

F=\frac{s^2_2}{s^2_1}

Solution to the problem  

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: \sigma^2_1 = \sigma^2_2

H1: \sigma^2_1 \new \sigma^2_2

a) Calculate the statistic

Now we can calculate the statistic like this:

F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have n_1 -1 =10-1=9 and for the denominator we have n_2 -1 =10-1=9 and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

p_v =2*P(F_{9,9}>1.727)=0.428

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be \alpha/2 =0.025, the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: F>4.026 \cup F since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the p_v > \alpha we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.  

8 0
3 years ago
Algebra 2! Please help!
slega [8]

B is the answer

As x → ∞, f(x) → - ∞

As x → - ∞, f(x) → - ∞


6 0
4 years ago
Read 2 more answers
If you are driving at 65 miles per hour how long would it take youth to drive 400 miles?
jeka57 [31]
It is going to be 6 hours and 15 mins
5 0
3 years ago
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