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lora16 [44]
3 years ago
9

A pair of sneakers cost the store $65. They mark them up by 40%. What is the selling price of the sneakers?

Mathematics
2 answers:
nadezda [96]3 years ago
5 0
The selling price is $91
Elina [12.6K]3 years ago
4 0

Answer:

91

Step-by-step explanation:

40% of 65 is 26. 26 + 65 = 91

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You develop an app to help students complete their homework. To earn money, you sell advertising space on the app’s main screen.
Oksanka [162]

Answer:

y = .003x + 25

Step-by-step explanation:

y = .003x + 25

3 0
4 years ago
Simplest form of a3b3c a-3b-3c-1
Serjik [45]

Answer:

a-6b-6c-2

Step-by-step explanation:

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3 years ago
PLEASE HELP QUICK!! Repost because typos | 100 POINTS
Oksi-84 [34.3K]

Answer:

Step-by-step explanation:

Well you have to first look at the chart to see if it fits what Ariel claimed.

There is actually a higher percent of students with no siblings that have pets if you look at the total for students with no siblings. If you look at the number though 45>39. Also her claim is not justified because the percentage is lower for students with siblings. Thanks for the 50 points ;) If you have any questions just ask

4 0
3 years ago
A coin contains 999 grams of nickel and 161616 grams of copper, for a total weight of 252525 grams.
Sedbober [7]

Answer:

X=64%

Step-by-step explanation:

The percentage of Cooper in the coin is 64%

How do you come to this conclusion?

252525--100%

161616--X%

Now you can use the cross rule: X=(161616x100)/252525

Check it out now X=64%

3 0
3 years ago
Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=10
Tju [1.3M]

Answer: 0.8238

Step-by-step explanation:

Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with \mu=106 and \sigma=15.

Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.

Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-

P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]

Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238

4 0
3 years ago
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