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JulsSmile [24]
3 years ago
15

During a​ one-month promotional​ campaign, Fran's Flix gave either a free DVD rental or a​ 12-serving box of microwave popcorn t

o new members. It cost the store ​$1 for each free rental and ​$2 for each box of popcorn. In​ all, 53 new members were signed up and the​ store's cost for the incentives was ​$94. How many of each incentive were given​ away?
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
8 0

Answer:

  • 12 free rentals
  • 41 boxes of popcorn

Step-by-step explanation:

If all of the new members had received free rentals, the cost would have been $53. The cost was actually ($94 -53) = $41 more than that. The additional cost of a box of popcorn is $1, so there must have been 41 boxes of popcorn in the mix of give-aways.

41 boxes of popcorn and 12 free DVD rentals were given away.

_____

<em>Check</em>

Cost of give-aways was $2×41 +$1×12 = $94, as required.

_____

If you like to write equations, you can let a variable represent the number of the highest-cost item (boxes of popcorn). If that is p, then the number of DVD rentals is (53-p) and the total cost is ...

  2p +(53 -p) = 94

  p = 94 -53 = 41 . . . . . . . boxes of popcorn

  (53 -p) = 53 -41 = 12 . . . DVD rentals

Hopefully, you can see the resemblance to the "word solution" above.

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You need to haul a load of patio bricks to a job site. Each brick weighs 4 pounds 14 ounces. Your truck can carry a 3/4 -ton loa
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You should be able to fit 370 in there.

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3 years ago
A marketing executive is studying Internet habits of married men and women during the 8am – 10pm time period on weekends ("prime
jolli1 [7]

Answer: Our required probability is 0.18.

Step-by-step explanation:

Since we have given that

Probability that husband is on the internet = 10% = 0.10

Probability that husband is not on the internet = 1-0.10 = 0.9

Probability that wife is on internet given that husband is on internet = 40% = 0.40

Probability that wife is on internet given that husband is not on internet = 20% = 0.20

Probability that wife is on internet is given by

0.1\times 0.4+0.9\times 0.2\\\\=0.04+0.18\\\\=0.22

So, Probability that the husband is also on internet given that wife is on internet is given by

\dfrac{P(\text{wife and husband both on internet)}}{P(wife\ on \ internet)}\\\\=\dfrac{0.4\times 0.1}{0.22}\\\\=\dfrac{0.04}{0.22}\\\\=0.18

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5 0
4 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
4 years ago
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