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Rudiy27
3 years ago
5

PLEASE HELP ME WITH THIS MATH QUESTION

Mathematics
1 answer:
Helen [10]3 years ago
5 0

Answer: 24%

Step-by-step explanation:

2610+8120 = The undergraduates and graduates combined.

That is 10730. You are figuring out the probability the student is a graduate when those two graduates are combined, because that is all the data given. So you would do 2610/10730 in your calculator, resulting in 24.324324324%. As it says rounded to the nearest percent in parentheses, it has to round to the whole number, 24%, and .3 rounds down.

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-6/13 - (-7/15)<br> pls solve it for me !!! :)
MAXImum [283]

Answer:

First, make the double negative a positive.

So, (-6/13) + (7/15)

Then multiply the numerator, 6×7=42

And denominator 13×15=195

42/195 simplify by dividing both by 3 so 14/65

8 0
2 years ago
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A car gets 27 mile per gallon of gas. How far will it go on 12 gallons?​
olga55 [171]

Answer:

324

Step-by-step explanation:

4 0
3 years ago
This question is for ktishler1 what is 1+1?
WARRIOR [948]

Answer: 11

if u put 1 and 1 together

u get 11

8 0
3 years ago
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Janet Fog's uncle, Pete Moore, promised her a gift of $30,000 upon her graduation from law school or $1500 every quarter for the
LenaWriter [7]
First calculate the future value of the annuity
The formula to find the future value of an annuity ordinary is
Fv=pmt [((1+r/k)^(kn)-1)÷(r/k)]
Fv future value?
PMT quarterly payment 1500
R interest rate 0.12
K compounded quarterly 4
N time 4 years
Fv=1,500×(((1+0.12÷4)^(4×4)
−1)÷(0.12÷4))
=30,235.32

Now compare the amount of the annuity with amount of the gift
30,235.32−30,000=235.32
So as you can see the amount of the annuity is better than the amount of the gift by 235.32

Second offer is better

Hope it helps!
5 0
3 years ago
Show that the equation x^3 + x -3 =0 has a solution between 1 and 2. That is the only information given
valentina_108 [34]

Answer:

Step-by-step explanation:

Given that:

f(x) = x^3 + x - 3 = 0

Since the given function is equal to zero, then the function :

f(x) = x^3 + x - 3

where;

x = 1 and

x = 2

f(x) = (1)^3 + (1) - 3 \\  \\ f(x) = 1 + 1 - 3 \\ \\f(x) = -1 < 0

f(2) =  x^3 + x - 3 \\ \\f(2) = (2)^3 + 3 - 3  \\ \\f(2) = 8 + 3 - 3  \\ \\f(2) = 11 - 3 \\ \\f(2) = 8 > 0 \\ \\

Thus; f(1) < 0 and f(2) > 0

8 0
2 years ago
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