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n200080 [17]
3 years ago
10

Billy and Sam started with the same number of baseball cards in their collections. Billy collected 3 cards per week and now has

29 cards. Sam collected 2 cards per week and now has 20 cards. Let x represent the number of cards they began with, and let y represnt the number of weeks. Which system of equations represents this situation?
Mathematics
2 answers:
IRISSAK [1]3 years ago
7 0
X+3y=29
x+27=20
These are the two equations for the given problem
Elina [12.6K]3 years ago
3 0

Answer:

x+3y=29

x+2y=20

Step-by-step explanation:

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A rectangular prism has a length of 4.5 cm, a width of 10 cm, and a height of 2.5 cm. What is the volume of the prism?
ruslelena [56]

Answer: 112.5 cm3

Hope that helped!

5 0
3 years ago
Question 8(Multiple Choice Worth 3 points)
Nastasia [14]
Answer: C) F(4) = 0

Detailed Explanation:

A zero/root of a polynomial is a value of the variable which makes the value of the polynomial to 0.

Therefore, when (x - 4) is a Factor :-
=> x - 4 = 0
=> x = 4

=> x = 4 is the Zero of the Polynomial.
4 0
2 years ago
Your hotel has 200 rooms in Dallas. There is a sport event coming up, and fans either book a rooms months before or wait for the
dedylja [7]

Answer:

89 rooms should be set for early book customer

Step-by-step explanation:

According to the given data we have the following:

OVERAGE(CO) = 200

SHORTAGE(CS) = 500

In order to calculate how many rooms should be set for early book customer we would have to use the following formula:

OPTIMAL BOOKING = MEAN + (Z * STDEV)

MEAN = 75

STDEV = 25

SERVICE LEVEL= CS / (CS + CO) = 500 / (500 + 200) = 0.7143

Z VALUE FOR 0.7143 = 0.57

OPTIMAL BOOKING =  75 + (0.57 * 25) = 89

89 rooms should be set for early book customer

3 0
3 years ago
F(x)=9x+5, find f^-1(x).
lidiya [134]

Answer:

interchange the role of x and y

x=9y+5

x-5=9y

<u>x-5</u><u> </u>=y

9

so,f^-1(x)= <u>x-5</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>9

hope it helps.

<h3>stay safe healthy and happy.</h3>
4 0
3 years ago
One household is to be selected at random from a town. ​ ​The probability that ​the household has a cat is 0.20.2 . ​ ​The proba
dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a ​cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

A = a + (A \cap B)

In which a is the probability that a household has a cat but not a dog and A \cap B is the probability that a household has both a cat and a dog.

By the same logic, we have that:

B = b + (A \cap B)

The probability that the household has a cat or a dog is 0.5

a + b + (A \cap B) = 0.5

The probability that the household has a dog ​is 0.4

B = 0.4

B = b + (A \cap B)

b = 0.4 - (A \cap B)

The probability that ​the household has a cat is 0.2.

A = 0.2

A = a + (A \cap B)

a = 0.2 - (A \cap B)

So

a + b + (A \cap B) = 0.5

0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5

A \cap B = 0.1

What is ​the probability that the household has a dog, given that the household has a ​cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5

There is a 50% probability that the household has a dog, given that the household has a ​cat.

4 0
3 years ago
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