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zepelin [54]
3 years ago
13

Can anyone tell me how to do 28% of 32

Mathematics
2 answers:
Korvikt [17]3 years ago
8 0
Well you do 32 divided by 28% and get 8.96
Arisa [49]3 years ago
7 0
28% of 32 would be added so if you want your answer it would be 32.28

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X=−4a 2 +3ab+12b 2 Y=9a 2 +ab−7b 2 ​ Y+X=Y+X=Y, plus, X, equals
dimaraw [331]

Answer:

5a  

2

+4ab+5b  

2

Step-by-step explanation:

step  1. (((0-(4•(a2)))+3ab)+(12•(b2)))+(((9•(a2))+ab)-7b2)

 step 2. (((0-(4•(a2)))+3ab)+(12•(b2)))+((32a2+ab)-7b2)

step 3.(((0-(4•(a2)))+3ab)+(22•3b2))+(9a2+ab-7b2)

step 4. (((0-(4•(a2)))+3ab)+(22•3b2))+(9a2+ab-7b2)

step 5. 5.1    Factoring    5a2 + 4ab + 5b2

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3 years ago
Use the associative property to simplify the expression.
BaLLatris [955]

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through

3 0
2 years ago
Read 2 more answers
What is the value of a in the equation? 22= a/11
OLga [1]

Answer:

Step-by-step explanation:

If you divide 33 by 11 it’ll be 3 so we can cross B off, and A too because it’s not a double digit

So that leaves us with C & D.

If we divided 222 and 11 it would leave us with 10 which doesn’t match 11, so that leaves us with D

(Sorry if I didn’t’t explain right )

6 0
3 years ago
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Find the value of a and b in (a,2)=(2,b)​
Ksenya-84 [330]

Answer:

a=2 and b=2...............

8 0
3 years ago
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
3 years ago
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