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enyata [817]
3 years ago
13

you are at the mall and you want to buy a hoodie for 35 dollars its on sale for 25 percent off how do you find the discounted pr

ice show all your work
Mathematics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:$8.75

Step-by-step explanation:Thus, a product that normally costs $35 with a 25 percent discount will cost you $26.25, and you saved $8.75. You can also calculate how much you save by simply moving the period in 25.00 percent two spaces to the left, and then multiply the result by $35 as follows: $35 x .25 = $8.75 savings.

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You would multiply 9x___ whatever the other number is witch is 4
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Make a equation with a slope of 2 and a y-intercept of -1
Nesterboy [21]

Answer:

y=2x-1

Step-by-step explanation:

y=mx+b where m=slope and b=y-intercept

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• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
3 years ago
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I don’t know the answer to this question
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27w^2-12 how do factor this in special cases
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Answer:

Take out the greatest common factor

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3(9w^2-4)

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