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Afina-wow [57]
3 years ago
11

An online streaming service deducts $7 from Serena’s account every month of a year. After 1 year, how much do the streaming serv

ice fees change Serena’s account?
Mathematics
2 answers:
V125BC [204]3 years ago
8 0
I think it's $84. Because there are 12 months in a year and they take $7 per month.
torisob [31]3 years ago
5 0

Answer:

$84

Step-by-step explanation:

This is a super simple question. All you have to know is how much money will be charged a month, and how many months there are in a year. Just simply multiply 7 by 12 as the streaming service is charging 7 dollars a month and there are 12 months in a year. After you multiply the two numbers, you should end up with 84 dollars charged after 1 year.

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5/6 and 1/6 are exponents to X. The X’s are divided by each other. does that make sense?
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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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