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2 years ago

15

Police estimate that 25% of drivers drive without their seat belts. If they stop 6 drivers at random, find theprobability that m

ore than 4 are wearing their seat belts.

1 answer:

earnstyle [38]2 years ago

8 0

Answer:

%17.80

Step-by-step explanation:

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What is 4/20 + 11/20 in its simplest form

vivado [14]

Answer:

Step-by-step explanation:

4/20 + 11/20 = 15/20

15/20 = 3/4 or 0.75

6 0

2 years ago

Read 2 more answers

Asanji spent half of his weekly allowance at the movies to earn more money his parents let him wash the car for $8 what is his w

Vera_Pavlovna [14]

$20 dollars if you start with $18 and subtract what his parents gave him its $10 then double it so $20 is the answer

4 0

2 years ago

Help asap pls -=-=-=-=--------------

bixtya [17]

Answer:

-1/3

Step-by-step explanation:

at x=0 y= 5

at x=3 y= 4

y=ax+b

b= 5

4=3x+5

3x= -1

x= -1/3

$y = - \frac{x}{3} + 5$

5 0

2 years ago

Given: PS=RT, PQ=ST<br> Prove: QS=RS

ivanzaharov [21]

Answer:

I) Eq(1) reason: sum of segments of a straight line

II) Eq(2) reason: Given PQ = ST & PS = RT

III) Eq(3) reason: sum of segments of a straight line

IV) Eq(4) reason: Same value on right hand sides of eq(2) and eq(3) demands that we must equate their respective left hand sides

V) Eq(5) reason: Usage of collection of like terms and subtraction provided this equation.

Step-by-step explanation:

We are given that;

PS = RT and that PQ = ST

Now, we want to prove that QS = RS.

From the diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

PQ + QS = PS - - - - (eq 1)

Now, from earlier we saw that PQ = ST & PS = RT

Thus putting ST for PQ & PS for RT in eq 1,we have;

ST + QS = RT - - - - (eq 2)

Again, from the line diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

RS + ST = RT - - - - -(eq 3)

From eq(2) & eq(3) we can see that both left hand sides is equal to RT.

Thus, we can equate both left hand sides with each other to give;

ST + QS = RS + ST - - - (eq 4)

Subtracting ST from both sides gives;

ST - ST + QS = RS + ST - ST

This gives;

QS = RS - - - - (eq 5)

Thus;

QS = RS

Proved

5 0

2 years ago

A researcher is interested in determining if the more than two thirds of students would support making the Tuesday before Thanks

klio [65]

Answer:

a. p = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are in favor of making the Tuesday before Thanksgiving a holiday, or they are against. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

So, the binomial probability distribution has two parameters, n and p.

In this problem, we have that $n = 1000$ and $p = \frac{700}{1000} = 0.7$ . So the parameter is

a. p = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.

8 0

2 years ago