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SIZIF [17.4K]
2 years ago
10

Based on an online poll, 35% of motorists routinely use their cell phone while driving. Tree people are chosen at random from a

group of 100. A) What is the probability of at least two people of the three people use their cell phone while driving? B) What is the probability that no more than one person of the three people use their cell phone while driving?
Mathematics
1 answer:
mart [117]2 years ago
7 0

Answer:

The anwerss to the question are

(A) P(No less than two people use their phones while driving) =  0.1225

(B) P(The probability that no more than one person of the three people use their cell phone while driving) = 0.147875

Step-by-step explanation:

The given relations are

Percentage of motorists that routinely drive while sing their phone = 35 %

The probaboloty that if a peerson is random;ty  selected from a group of hudred person routinely uses their phone wjile friving P(phone) = 35

The probability that a  motorist randomly selected fron a set of 100 do not routinely use thir phones while driving = P(No celll phone)  = 65

Then the probability that when three people are selected at random at least two people of the three people use their cell phone while driving is

P(phone)  = 35/100m = 0.35

P(No celll phone)  = 65/100 = 0.65

(A) Probability of at least two of three use their phones whle driving is

0.35×0.35×0.65 +0.35×0.35×0.35 = 0.1225

(B) The probability of only one person out of three seted use their phones while driving is

(0.35)(0.65)(0.65) = 0.147875

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2 years ago
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Peter, Gordon and Gavin share £36 in a ratio 2:1:1. How much money does each person get?
melamori03 [73]

Answer: let the common multiple be x..

according to given condition,

2x + x +x =36

4x = 9

Peter = 2x = 2 ×9 = £18

Gordon = Gavin = x = £9

Hope this helps have a amazing day bye!

8 0
1 year ago
Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 1
BigorU [14]

Answer:

a:  Sample mean: 20, Sample standard deviation: 1.732

b:  19.13 <µ < 20.87

c:  18.84< µ < 21.16

d:  as the confidence level increases, the interval becomes wider

Step-by-step explanation:

a:  Sample mean is found by adding up all the individual values of the sample, then dividing by the total number of values in the sample.  

Sample standard deviation is the square root of the variance

See the first attached photo for the calculations of these values.

We need to create a 80% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.20 (80%), and row 7 (degree of freedom)

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See the second attached photo for the construction of the confidence interval

We need to create a 90% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.10 (90%), and row 7 (degree of freedom)

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3 0
2 years ago
A middle school chess club has 5 members: Adam, Bradley, Carol, Dave, and Ella. Two students from the club will be selected at r
laila [671]

Answer:

Probability that Adam and Ella will be selected:

\displaystyle \frac{1}{10}=0.1

Step-by-step explanation:

Probabilities

The probability of a random event E to occur is a real number between 0 and 1, both inclusive, where 0 indicates an impossible event and 1 a sure event. There are many techniques to compute probabilities depending on the particular situation and distribution.

This question will be solved by simple calculations and logic, given its simplicity. We know the middle school chess club has 5 members: Adam, Bradley, Carol, Dave, and Ella. Two of them are going to be selected at random to participate in the county chess tournament. We can calculate the number of different ways it can be done without any restriction. It's called the sample space.

The sample space of this event is the combination of 5 members regardless of their position. If {a,b,c,d,e} are the five members, then the possible combinations are {ab,ac,ad,ae,bc,bd,be,cd,ce,de}. Notice that there are only 10 possibilities because the combination ab is the same as ba since it's the same team for the tournament.

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3 0
2 years ago
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mixas84 [53]
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47.6 - 41.5 = 6.1

I don't know about the cubic centimeters part, I don't know for sure if mL and cubic cm are the same, sorry. :(
8 0
2 years ago
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