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Delvig [45]
3 years ago
12

Christais attending the county fair that charges a $12 entry fee. The entry fee includes 10 free rides, but any additional rides

cost $1.50 each.
Write a function that models Christa's county fair expenses, cr), when she rides ""rides.
Mathematics
1 answer:
ratelena [41]3 years ago
8 0

Answer:

he saves 3 dollars. its not worth it.

Step-by-step explanation:

you divide 12 by ten getting you 1.2 which you subtract from 1.50 that then multiply that by 10

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QUESTION 58 Kendra had a choice of 3 breads, 3 meats, and 2 cheeses to make a sandwich. How many different sandwiches can she ma
LUCKY_DIMON [66]

Assume 3 breads are all different, 3 meats are all different and 2 cheeses are all different. A sandwich contains a type of bread, a type of meat and a type of cheese. The number of different types of sandwiches that can be made is

3*3*2=18

The correct choice is (C).

7 0
3 years ago
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Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are sel
serious [3.7K]

Complete Questions:

Find the probability of selecting none of the correct six integers in a lottery, where the order in which these integers are selected does not matter, from the positive integers not exceeding the given integers.

a. 40

b. 48

c. 56

d. 64

Answer:

a. 0.35

b. 0.43

c. 0.49

d. 0.54

Step-by-step explanation:

(a)

The objective is to find the probability of selecting none of the correct six integers from the positive integers not exceeding 40.  

Let s be the sample space of all integer not exceeding 40.

The total number of ways to select 6 numbers from 40 is |S| = C(40,6).

Let E be the event of selecting none of the correct six integers.

The total number of ways to select the 6 incorrect numbers from 34 numbers is:

|E| = C(34,6)

Thus, the probability of selecting none of the correct six integers, when the order in which they are selected does rot matter is  

P(E) = \frac{|E|}{|S|}

         = \frac{C(34, 6)}{C(40, 6)}\\\\= \frac{1344904}{3838380}\\\\=0.35

Therefore, the probability is 0.35

Check the attached files for additionals  

8 0
3 years ago
Consider three events A, B and C with following properties.
lisov135 [29]
By the inclusion/exclusion principle,

\mathbb P(D)=\mathbb P(A\cup B\cup C)
\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)
\mathbb P(D)=\dfrac14+\dfrac16+\dfrac14-\dfrac39+\dfrac1{13}
\mathbb P(D)=\dfrac{16}{39}\neq1

So the union of A, B, and C does not constitute the entire sample space.
5 0
4 years ago
The diagram shows a square divided into strips of equal width. Three strips are black and two are grey. What fraction of the per
astra-53 [7]
2/5 at the top nd bottom, but 0/5 on the sides. So 2/5 + 0/5= 1/5
8 0
3 years ago
VERY EASY HALP PLEASE
Nina [5.8K]
The second option ! 2.54 - 2.3 = 0.24
6 0
3 years ago
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