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wolverine [178]
3 years ago
9

Party favors are on sale for $2.50 each. You have $380 to spend on the decorations and gifts, and you have already spent $272 on

decorations. How many party favors could you buy.
Mathematics
2 answers:
Artyom0805 [142]3 years ago
8 0

Answer: 39

Step-by-step explanation:

You would subtract 380-272 to get 108 then multiple 2.50 to 39 to get $97.50 plus tax will take the rest of the money.

Alenkinab [10]3 years ago
3 0

Answer:

we could buy 43 party favors

Step-by-step explanation:

initially we have an amount of $380

if we already spend an amount of $272 this we have to subtract it from the total

$380 - $272 = $108

if each party favor comes out $ 2.50 and we have $ 108 we have to divide what we have by what each one comes out to know how many we can buy

$108 / $2.50 = 43.2

this means we could buy 43 party favors

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Ksivusya [100]

Answer:

  • 2 1/3 minutes

Step-by-step explanation:

<u>Distance:</u>

  • d = 7/8 mile

<u>Speed:</u>

  • s = 3/8 mpm

<u>Time:</u>

  • t = d/s
  • t = 7/8 : 3/8 = 7/8 * 8/3 = 7/3 = 2 1/3 min
8 0
2 years ago
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\Delta = \frac{1}{2}\left|\begin{array}{ccc}1&0&0\\1&6&2\\1&4&8\end{array}\right |\\&#10;\Delta = 0.5(48-8)\\&#10;\Delta = 20 \;square\; units


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Find the lateral area of the cone in terms of pi.
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\pi r\sqrt{h^2+r^2}

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8 0
3 years ago
A consumer organization estimates that over a​ 1-year period 20​% of cars will need to be repaired​ once, 5​% will need repairs​
gtnhenbr [62]

Answer:

a) 0.5476

b) 0.0676

c) 0.4524

Step-by-step explanation:

Given this information, we can conclude that 74% of the cars won't need any repairs over a 1-year period (100 - 20 - 5 - 1 = 74%). And 26% will need at least 1 repair over a 1-year period.

P(car doesn't need repair) = 0.74

P (car needs repair) = 0.26

If you own two cars, the probability that:

<u>a) Neither will need repair:</u>

We need that car 1 won't need repair AND car 2 won't need repair.

=P(Car 1 doesn't need repair) x P(Car 2 doesn't need repair)

= 0.74 x 0.74 = 0.5476

The probability that neither will need repair is 0.5476.

<u>b) Both will need repair:</u>

We need that car 1 needs repair AND car 2 needs repair.

P(Car 1 needs repair) x P(Car 2 needs repair)

= 0.26 x 0.26 = 0.0676

The probability that both will need repair is 0.0676

<u>c) At least one car will need repair</u>

Car 1 needs repair or Car 2 needs repair or both need repair.

To solve this one, it's easier to use the complement of P(neither needs repair)

1 - P(neither needs repair)

1 - (0.74)(0.74)  = 1 - 0.5476 = 0.4524

The probability that at least one car will need repair is 0.4524

4 0
3 years ago
Solve the following systems by substitution
Paladinen [302]

Answer:

The answer to your question is below

Step-by-step explanation:

Question 1

                x = 5                       Equation l

                2x + y = 10              Equation ll

- Substitute Equation l in equation ll

               2(5) + y = 10

                y = 10 - 10

                y = 0

- Solution  (5, 0)

Question 2

                 x + 16y = 20          Equation l

                 x = 4y                    Equation ll

Substitute equation ll in equation l

                4y + 16y = 20

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                           y = 20/20

                           y = 1

-Find x

                x = 4(1)

                x = 4

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                (4, 1)

Question 3

                   2x + 8y = 20            Equation l

                     x = 2                       Equation ll

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                            8y = 16

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- Solution

                   (2, 2)              

8 0
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