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muminat
3 years ago
15

I need to know how many cupcakes are needed to fill this box.

Mathematics
1 answer:
Gre4nikov [31]3 years ago
4 0

Is it more to this question like numbers or something I don’t really understand
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Distributed property for 0.75(3.5a-6b)
Alika [10]
0.75(3.5a -6b)
= 0.75* (3.5a) -0.75* (6b) (distributive property)
= 2.625a -4.5b

The final answer is 2.625a -4.5b~
5 0
3 years ago
Find the circumference of this circle.
finlep [7]

Answer:

81.64

Step-by-step explanation:

2(3.14)=6.28

6.28*13=81.64

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3 0
3 years ago
A bookstores sells frequent buyer Discount cards For $12 each. The cost of each book with the discount card is $7. Cost of a boo
Brilliant_brown [7]

Answer:

6 book purchases.

Step-by-step explanation:

As the bookstore sells frequent buyers discount cards at $ 12, which make the books that usually cost $ 9 cost $ 7, to determine from how many books a card-holder and a common buyer will spend the same amount of money is necessary perform the following calculation:

9 - 7 = 2

12/2 = 6

Therefore, after purchasing 6 books, both categories of buyers will have spent the same amount of money. This is verified with the following calculation:

Card holder = 12 + 7 x 6 = 54

Non-card holder = 9 x 6 = 54

8 0
3 years ago
What two numbers is 94 between
levacccp [35]
94 is between 93 and 95
7 0
4 years ago
Read 2 more answers
A medical device company knows that 11% of patients experience injection-site reactions with the current needle. If 4 people rec
Orlov [11]

Answer:  0.6274

Step-by-step explanation:

Given: The proportion of patients experience injection-site reactions with the current needle : p=0.11

Sample size : n= 4

Let x be a binomial random variable that represents the people get an injection-site reaction.

Binomial probability formula: P(X=x)= ^nC_x p^x(1-p)^{n-x}

The required probability : P(x=0)

=\ ^4C_0(0.11)^0(1-0.11)^4\\\\=(1)(1)(0.89)^4\\\\=0.62742241\\\approx0.6274

Hence, the  probability that none of the 4 people get an injection-site reaction =  0.6274

6 0
4 years ago
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