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frutty [35]
3 years ago
12

What type of graph is best to use to show data that are parts of a whole?

Mathematics
1 answer:
san4es73 [151]3 years ago
6 0
A pie chart? i think
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A supermarket sells milk in two sizes: 4-pint
Lesechka [4]

Find the unit rate by dividing price by total units:

0.98 / 4 pints = 0.245 per pint

1.46/ 6 pints = 0.243 per pint

The 6 pint size has a lower price per point so it is the better deal.

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A company charges $211.25 for 5 trees and 15 shrubs. The company charges 15.25 more for a tree than a shrub. How much does each
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The answer is $1.96 I hope I could help?
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What sentence about markdowns and markups is true?
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The answer is C.
<span>C. A markup has no limit, but a markdown is limited to 100% or less. </span>
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Solve for x<br><br> x/4= 25<br><br> quick
storchak [24]

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Step-by-step explanation:

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There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit
DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = nCr * p^{r} * q^{n-r}

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = 15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}

p(x=3) = 455 * 0.4^{3} * 0.6^{12}

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday  = 1 - 0.0633 = 0.9337

3 0
3 years ago
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