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natulia [17]
2 years ago
5

drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This mon

th, the insurance company writes 4 new policies for adults earning their first drivers license. What is the probability that these 4 will contain at least two more high-risk drivers than low-risk drivers
Mathematics
1 answer:
Kisachek [45]2 years ago
5 0

Complete question is:

A large pool of adults earning their first drivers license includes 50% low- risk drivers, 30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This month, the insurance company writes 4 new policies for adults earning their first drivers license. What is the probability that these 4 will contain at least two more high-risk drivers than low-risk drivers?

Answer:

probability that these 4 will contain at least two more high-risk drivers than low-risk drivers = 0.0488

Step-by-step explanation:

Let H represent High risk

M represent moderate risk

L represent Low risk.

The following combinations will satisfy the condition that there are at least two more high-risk drivers than low-risk drivers: HHHH, HHHL, HHHM, HHMM

The HHHH case has probability 0.2 ⁴ = 0.0016

The HHHL case has probability 4 × 0.2³ × 0.3 = 0.0096 (This is because L can be in four different places)

Similarly, the HHHM case has probability 4 × 0.2 ³ × 0.5 = 0.016

Lastly, the HHMM case has probability 6 × 0.2 ² × 0.3 ² = 0.0216 (This is because the number of ways to choose places for two M letters in this way is 6)

Summing all these probabilities, we have;

0.0016 + 0.0096 + 0.016 + 0.0216 = 0.0488

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If r and s are positive integers, is \small \frac{r}{s} an integer? (1) Every factor of s is also a factor of r. (2) Every prime
Yuri [45]

Answer:

<em>If statement(1) holds true, it is correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em>If statement(2) holds true, it is not necessarily correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em></em>

Step-by-step explanation:

Given two positive integers r and s.

To check whether \small \frac{r}{s} is an integer:

Condition (1):

Every factor of s is also a factor of r.

r \geq s

Let us consider an example:

s = 5^2 \cdot 2\\r = 5^3 \cdot 2^2

\dfrac{r}{s} = \dfrac{5^3\cdot2^2}{5^2\cdot2} = 10

which is an integer.

Actually, in this situation s is a factor of r.

Condition 2:

Every prime factor of <em>s</em> is also a prime factor of <em>r</em>.

(But the powers of prime factors need not be equal as we are not given the conditions related to powers of prime factors.)

Let

r = 2^2\cdot 5\\s =2^4\cdot 5

\dfrac{r}{s} = \dfrac{2^3\cdot5}{2^4\cdot5} = \dfrac{1}{2}

which is not an integer.

So, the answer is:

<em>If statement(1) holds true, it is correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em>If statement(2) holds true, it is not necessarily correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em></em>

8 0
3 years ago
75 POINTS. I WILL GIVE BRAINLIEST.
agasfer [191]

Question 5, x could be either 6 or -1.

3 0
3 years ago
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(5x-7)+(6+33)<br> solve the equation
Vinil7 [7]

Answer: 5x + 32

Step-by-step explanation:

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3 years ago
At Grayson's Pancake House there are 86 dirty coffee mugs and 774 clean coffee mugs. What percentage of the coffee mugs at the p
Marina86 [1]

Answer:

10%

Step-by-step explanation:

774+86=860

86/860=d/100

860d=8600

8600÷860=10

Hope this helps

7 0
3 years ago
Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:
tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

  • 91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

  • 79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

Q1 = \frac{N +1}{4}th

So, we have:

Q1 = \frac{13 +1}{4}th

Q1 = \frac{14}{4}th

Q1 = 3.5th

This is the average of the 3rd and the 4th element

Q1 = \frac{1}{2} \times (84 + 91)

Q1 = 87.5

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

Q3 =  3 \times \frac{N +1}{4}th

So, we have:

Q3 =  3 \times \frac{13 +1}{4}th

Q3 =  3 \times 3.5th

Q3 =  10.5th

This is the average of the 10th and the 11th element.

Q_3 =\frac12 \times (98 + 101)

Q_3 =99.5

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

IQR = Q_3 -Q_1

IQR = 99.5 - 87.5

IQR = 12

Also, we have:

IQR(1.5) = 12 \times 1.5

IQR(1.5) = 18

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

Lower = Q_1 - IQR(1.5)

Lower = 87.5- 18

Lower = 69.5

Upper = Q_3 + IQR(1.5)

Upper = 99.5 + 18

Upper = 117.5

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

6 0
2 years ago
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