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3 years ago

7

Convert to a decimal: 68⁄100 = (Do not round. Enter all decimal digits.)

2 answers:

Helen [10]3 years ago

8 0

Hello!

To convert a fraction to a decimal you have to make the denominator 100

Since the denominator is already 100 we put the numerator behind the decimal point

The answer is 0.68

Hope this helps!

Mandarinka [93]3 years ago

4 0

Hey there!

Here is your answer:

<em><u>The proper answer to this question is "0.68".</u></em>

Reason:

<u><em> $\frac{68}{100}$ = $\frac{17}{25}$ </em></u>

<span><u><em>68/100</em></u></span><u><em> = ( 68 ÷ 4 ) over ( 100 ÷ 4 ) </em></u><span><u><em> 68 ÷ 4 100 ÷ 4 </em></u></span><u><em> = 17 over 25 </em></u><span><u><em>1725</em></u></span><u><em> </em></u>

<em>Therefore the answer is "0.68".</em>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit

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Answer:

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b. $P(E_2/A)=0.395$ \

c. $P(E_3/A)=0.526$

Step-by-step explanation:

Let $E_1,E_2,E_3$ are the events that denotes the good drive, medium drive and poor risk driver.

$P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20$

Let A be the event that denotes an accident.

$P(A/E_1)=0.01$

$P(A/E_2=0.03$

$P(A/E_3)=0.10$

The company sells Mr. Brophyan insurance policy and he has an accident.

a.We have to find the probability Mr.Brophy is a good driver

Bayes theorem, $P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}$

We have to find $P(E_1/A)$

Using the Bayes theorem

$P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

Substitute the values then we get

$P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}$

$P(E_1/A)=0.0789$

b.We have to find the probability Mr.Brophy is a medium driver

$P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395$

c.We have to find the probability Mr.Brophy is a poor driver

$P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526$

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3 years ago

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Ymorist [56]

Answer:

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Step-by-step explanation:

Suppose we have two events, A and B, the conditional probability formula is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

In which

P(A|B) is the probability of A happening given that B happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(B) is the probability of B happening.

In this problem, we have that:

$P(A \cap B) = \frac{3}{8}, P(B) = \frac{5}{7}$

So

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3 years ago