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svet-max [94.6K]

1 year ago

5

A sample of 2,000 licensed drivers revealed the following number of speeding violations. Number of Violations Number of Drivers

0 1,910 1 46 2 18 3 12 4 9 5 or more 5 Total 2,000 Click here for the Excel Data File What is the probability that a particular driver had exactly two speeding violations

1 answer:

Arturiano [62]1 year ago

6 0

Answer:

The probability that a particular driver had exactly two speeding violations is 0.009.

Step-by-step explanation:

We are given that a sample of 2,000 licensed drivers revealed the following number of speeding violations;

<u>Number of Violations</u> <u>Number of Drivers</u>

0 1,910

1 46

2 18

3 12

4 9

5 or more <u> 5 </u>

<u>Total</u> <u> 2000 </u>

<u />

Now, the data means that 1,910 drivers had 0 speeding violations and so on.

Now, we have to find the probability that a particular driver had exactly two speeding violations, that means;

Number of drivers having exactly two speeding violations = 18

Total numbers of drivers = 2000

So, the required probability = $\frac{\text{No. of drivers having two speeding violations}}{\text{Total no. of drivers}}$

= $\frac{18}{2000}$ = <u>0.009</u>

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For which values of a the system has no solution: x≤5, x≥a

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Given:

The system of inequalities is

$x\leq 5$

$x\geq a$

To find:

The values of a for which the system has no solution.

Solution:

We have,

$x\leq 5$ ...(1)

It means the value of x is less than or equal to 5.

$x\geq a$ ...(2)

It means the value of x is greater than or equal to a

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$a\leq x\leq 5$

But if a is great than 5, then there is no value of which satisfies this inequality.

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2 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

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