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enot [183]
3 years ago
12

A consumer organization estimates that over a​ 1-year period 15​% of cars will need to be repaired​ once, 9​% will need repairs​

twice, and 4​% will require three or more repairs. If you own two​ cars, what is the probability that ​a) neither will need​ repair? ​b) both will need​ repair? ​c) at least one car will need​ repair?
Mathematics
1 answer:
Llana [10]3 years ago
5 0

Answer:

a) The probability that neither will need repair is 51.84%

b) The probability that both will need repair is 7.84%

c) The probability that at least one car will need repair is 56%

Step-by-step explanation:

a) The probability that neither will need repair is 51.84%

(100 – 15 – 9 – 4)/100 * (100 – 15 – 9 – 4)/100  

(72/100) * (72/100)

5,184 / 10,000 = 0.5184 in decimals and 51.84%  

b) The probability that both will need repair is 7.84%

(15 + 9 + 4)/100 * (15 + 9 + 4)/100  

(28/100) * (28/100)

784/ 10,000 = 0.0784 in decimals and 7.84%  

c) The probability that at least one car will need repair is 56%

(15 + 9 + 4)/100 + (15 + 9 + 4)/100  

(28/100) + (28/100)

56/ 100 = 0.56 in decimals and 56%  

To calculate probabilities of two or more events are two rules: the sum and the product. The sum rule is used to calculate the probability of two or more events that cannot occur at the same time or are dependent. For example, the probability of having a 2 or a 3 when you throw a six-sided dice is 33%.

P (2) + P (3) = P

1/6 + 1/6 = 2/6 = 1/3 in fraction; 0.33 in decimals; 33% in percentage

 

On the other hand, the product rule is used to calculate the probability of two or more event that can occur at the same time or are independent. For example, the probability of having a 2 in both dices when you throw two six-sided dices is 2.78%.

P (2 in first dice) x P (2 in second dice) = P

1/6 * 1/6 = 1/36 in fraction; 0.0278 in decimals; 2.78% in percentage

Finally, you can use both rules. For example, the probability of having a 2 or a 3 in both dices when you throw two six-sided dices is 2.78%.

P (in first dice) x P (in second dice) = P

(1/6 + 1/6) * (1/6 + 1/6) = P

(2/6) * (2/6) = P

(1/3) * (1/3) = P

P = 1/9 in fraction; 0.111 in decimals; 11.11% in percentage

 

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The annual tuition at a specific college was $20,500 in 2000, and $45,4120
nika2105 [10]

Answer: the tuition in 2020 is $502300

Step-by-step explanation:

The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.

The formula for determining the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

a represents the first term of the sequence.

d represents the common difference.

n represents the number of terms in the sequence.

From the information given,

a = $20500

The fee in 2018 is the 19th term of the sequence. Therefore,

T19 = $45,4120

n = 19

Therefore,

454120 = 20500 + (19 - 1) d

454120 - 20500 = 19d

18d = 433620

d = 24090

Therefore, an

equation that can be used to find the tuition y for x years after 2000 is

y = 20500 + 24090(x - 1)

Therefore, at 2020,

n = 21

y = 20500 + 24090(21 - 1)

y = 20500 + 481800

y = $502300

6 0
3 years ago
What is the percent increase between getting a high school scholarship and bachelors degree when high school scholarship is $421
gogolik [260]

Answer:

The percent increase between getting a high school scholarship and bachelor's degree is <u>59.14%</u>.

Step-by-step explanation:

Given:

High school scholarship is $421.

Bachelor's degree is $670.

Now, to find the percent increase between a high school scholarship and bachelor's degree.

So, we get the amount of increase between a high school scholarship and bachelor's degree.

670-421=249.

<em>Thus, the amount of increase = $249</em>.

Now, to get the percent increase between a high school scholarship and bachelor's degree:

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=59.14\%.

Therefore, the percent increase between getting a high school scholarship and bachelor's degree is 59.14%.

7 0
3 years ago
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