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o-na [289]
3 years ago
10

Nola was selling tickets at the high school dance. At the end of the evening, she picked up the cash box and noticed a dollar ly

ing on the floor next to it. She said, I wonder whether the dollar belongs inside the cash box or not. The price of tickets for the dance was 1 ticket for $5 (for individuals) or 2 tickets for $8 (for couples). She looked inside the cash box and found $200 and ticket stubs for the 47 students in attendance. Does the dollar belong inside the cash box or not? (justify/defend)
Mathematics
1 answer:
Black_prince [1.1K]3 years ago
5 0

Answer:

yes

Step-by-step explanation:

s + 2c = 47

5s+8c = 200

which one checks as in the other solution does not have a solution in non-negative integers. On the other hand, the analogous system

s + 2c = 47

5s+8c = 201

with solution s=13 and c=17. So, as in the other solution, we conclude that the extra dollar belongs in the cash box, with 13 singles and 17 couples attending the dance.

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What is this supposed to be 5-8=?
erica [24]
Answer is 3 so simple
8 0
2 years ago
Amy buys two shirts. One shirt costs $12.95 and the other costs $15.98. How much money does Amy spend altogether
shtirl [24]

Answer:

$28.93

Step-by-step explanation:

 12.95+15.98 = $28.93

4 0
3 years ago
Read 2 more answers
A company manager for a tire manufacturer is in charge of making sure there is the least amount of defective tires. If the tire
Anestetic [448]

Answer:

0.347% of the total tires will be rejected as underweight.

Step-by-step explanation:

For a standard normal distribution, (with mean 0 and standard deviation 1), the lower and upper quartiles are located at -0.67448 and +0.67448 respectively. Thus the interquartile range (IQR) is 1.34896.

And the manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires.

1.5 of the Interquartile range = 1.5 × 1.34896 = 2.02344

1.5 of the interquartile range below the lower quartile = (lower quartile) - (1.5 of Interquartile range) = -0.67448 - 2.02344 = -2.69792

The proportion of tires that will fall 1.5 of the interquartile range below the lower quartile = P(x < -2.69792) ≈ P(x < -2.70)

Using data from the normal distribution table

P(x < -2.70) = 0.00347 = 0.347% of the total tires will be rejected as underweight

Hope this Helps!!!

6 0
4 years ago
The cost in dollars of making x items is given by the function C(x)=10x+900 .
Stella [2.4K]

For the given cost equation we have:

a) Fixed cost is $900.

b) For making 25 items the cost is $1150.

c) D: x ∈ [0, 150]

   R: c ∈ [900, 2400]

<h3>Working with the cost equation.</h3>

Here we know that the cost equation is:

c(x) = 10*x + 900.

First, we want to get the fixed cost, it is given by evaluating the function in x = 0.

c(0) = 10*0 + 900 = 900

The fixed cost is 900.

b) Now we want to get the cost for making 25 items, to get this, we just evaluate in x = 25.

c(25) = 10*25 + 900 = 250 + 900 = 1150

c) Now, if the maximum cost is 2400, then the maximum number of items that we can make is x₀, such that:

c( x₀) = 2400 = 10*x₀ + 900

Solving for x₀ we get:

x₀ = (2400 - 900)/10 = 150

Now we want to get the range and domain.

We know that we can make between 0 and 150 items, so the domain is:

D: x ∈ [0, 150]

For the range, we know that the fixed cost for 0 items is 900, and the maximum cost is 2400, then the range is:

R: c ∈ [900, 2400]

If you want to learn more about domain and range:

brainly.com/question/10197594

#SPJ1

5 0
2 years ago
The lengths of the sides of a triangle are 5, 12, and 13. What is the length of the altitude drawn to the side with length equal
mario62 [17]

Answer:  using  A= bh/2   THe height is 9.23

Step-by-step explanation:  First, with the Right Angle at the bottom, usr the sides to compute the area:  12*5=60

THen Imagine the side=13 as the base, so you have  b=13 for the formula

use the formula A= bh/2

60 = 13h/2 ==> 2(60) =13h --> 120/13 = h

h = 9.23

5 0
3 years ago
Read 2 more answers
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