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enot [183]
3 years ago
13

A prime number is a number that is evenly divisible only by 1 and itself. The prime numbers less than 100 are listed below.2 3 5

7 11 13 17 19 23 29 3137 41 43 47 53 59 61 67 71 73 7983 89 97Choose one of these numbers at random. Find the probability thata. The number is odd b. The sum of the digits is odd c. The number is greater than 70
Mathematics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

a. 0.96 = 96% probability that the number is odd.

b. 0.48 = 48% probability that the sum of the digits is odd.

c. 0.24 = 24% probability that the number is greater than 70.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

We are given the prime numbers less than 100, they are:

2,3,5,7,11,13,17,19,23,29,31,37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

There are 25 prime numbers less than 100.

a. The number is odd

Of the 25, 24 are odd, only the number 2 is even. So

24/25 = 0.96 = 96%

0.96 = 96% probability that the number is odd.

b. The sum of the digits is odd

The sum of the digits is odd for:

3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89. So for 12 of them

12/25 = 0.48 = 48%

0.48 = 48% probability that the sum of the digits is odd.

c. The number is greater than 70

71, 73, 79, 83, 89, 97. 6 of them

6/25 = 0.24 = 24%

0.24 = 24% probability that the number is greater than 70.

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Connor is going to see a movie and is taking his 4 kids. Each movie ticket costs $13 and there are an assortment of snacks avail
EleoNora [17]

If Connor buys 3 snacks at 3 dollars each for himself and his 3 children, the net cost (including movie tickets) would be 88 dollars.

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Step-by-step explanation:

To find the cost of the snacks:

Cost = Price*NumberOfSnacks, or C=p*n, or C=pn

To find the unit cost of the snack, find the price, the p value, which is 3 dollars.

Then, find the total number of snacks that will be bought. Since each person receives an equal number of snacks, 3 per person, and there are 4 people (himself and his 3 children), we can multiply 3*4 to find the n value, which is now known to be 12.

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To find the cost if he buys x amount of snacks for himself and his 3 children (assumed to be "everyone"), all costs but the n value are to be the same, creating this equation:

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2 years ago
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To calculate the combination of 2 students to be chosen, we use:

\begin{gathered} ^4C_2=\frac{n!}{(n-r)!}=\frac{4!}{(4-2)!}=\frac{4!}{2!} \\ ^4C_2=\frac{4\cdot3\cdot2\cdot1}{2\cdot1}=\frac{4\cdot3}{1}=12 \\ ^4C_2=12 \end{gathered}

Therefore, there 12 possible combinations from these

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1 year ago
(PLEASE HELP!!)
AlexFokin [52]

Answer:

X= 8

Y= 130

(8, 130)

Step-by-step explanation:

I used a graphing calculator to see where the two equations intersected.

Hope this helped :)

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3 years ago
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