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1 year ago

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Trevor Ariza is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next on

e, and he makes his free throws 70%, percent of the time. What is the probability of Trevor Ariza making none of his next 5 free throw attempts?

2 answers:

vivado [14]1 year ago

7 0

Answer:

Probability of making none of next 5 free throws = 0.00243

Step-by-step explanation:

"Making or missing free throws doesn't change the probability that he will make his next one"

So the probability of making none of next 5 free throws

= Probability of missing next 5 free throws

= Probability of missing 1 free throw^5

"...he makes his free throws 70%" so the probability of missing 1 free throw = 1-70% = 0.3

So the probability of making none of next 5 free throws = 0.3^5

= 0.00243

zzz [600]1 year ago

7 0

Answer:

0.0024

Step-by-step explanation:

Making or missing free throws doesn't change the probability of next FTs

P(none of next 5 FTs)

=P(miss next FT)*P(miss next FT)*P(miss next FT)*P(miss next FT)*P(miss next FT)

=P(miss next FT)^5

= (1 - P(make next FT))^5

= (1 - 0.7)^5

= 0.3^5

=0.0024

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The ratio of money in Brian's wallet to Colin's wallet one day was 5:1. Brian spent £27 that day. Brian now had £3 less than Col

myrzilka [38]

Answer:

Brian had $30 initially.Brian had $30 initially.

Step-by-step explanation:

WE are given the following in the question:

Let Brian have x dollars and Colin have y dollars.

Ration of Brian's money to Colin's money is 5:1. This, we can write the equation:

$\dfrac{x}{y} = \dfrac{5}{1}\\\\x = 5y$

"Brian spent £27 that day. Brian now had £3 less than Colin."

Thus, we can write the equation:

$x-27 = y-3\\x-y = 24$

Solving the two equation by substitution, we get,

$x - y - 24\\5y - y = 24\\4y = 24\\y=6\\x = 5y =30$

Thus, Brian had $30 initially.

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1 year ago

Alicia borrowed $15,000 to buy a car. She borrowed the money at 8% for 6 years. What is the interest she will pay for the loan ?

kykrilka [37]

The interest rate she would pay would be is $7,200

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1 year ago

The number of tickets purchased by an individual for Beckham College's holiday music festival is a uniformly distributed random

egoroff_w [7]

Answer:

The mean is 4.5 and the standard deviation is 1.44.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform probability distribution is:

$M = \frac{(a + b)}{2}$

The standard deviation of the uniform probability distribution is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

Uniformly distributed random variable ranging from 2 to 7.

This means that $a = 2, b = 7$ .

So

$M = \frac{(2 + 7)}{2} = 4.5$

$S = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(7 - 2)^{2}}{12}} = 1.44$

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1 year ago

Thomas graphed the line that represents the equation y=34x.

zloy xaker [14]

Answer:

The ordered pairs represent points on the line are

(4, 3) ⇒ C

(2, $\frac{3}{2}$ ) ⇒ D

(-8, -6) ⇒ E

Step-by-step explanation:

To find the ordered pairs represent points on the line, substitute x by the x-coordinate of each point, if the value of y equals the y-coordinate of the point, then the point is on the line.

∵ The equation is y = $\frac{3}{4}$ x

∵ The ordered pair is (8, $\frac{1}{6}$ )

→ Substitute x by 8

∴ y = $\frac{3}{4}$ (8)

∴ y = 6

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair (8, $\frac{1}{6}$ ) does not represent a point on the line

∵ The ordered pair is ( $\frac{-2}{3}$ , $\frac{1}{2}$ )

→ Substitute x by $\frac{-2}{3}$

∴ y = $\frac{3}{4}$ ( $\frac{-2}{3}$ )

∴ y = $\frac{-1}{2}$

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair ( $\frac{-2}{3}$ , $\frac{1}{2}$ ) does not represent a point on the line

∵ The ordered pair is (4, 3 )

→ Substitute x by 4

∴ y = $\frac{3}{4}$ (4)

∴ y = 3

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (4, 3) represents a point on the line

∵ The ordered pair is (2, $\frac{3}{2}$ )

→ Substitute x by 2

∴ y = $\frac{3}{4}$ (2)

∴ y = $\frac{3}{2}$

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (2, $\frac{3}{2}$ ) represents a point on the line

∵ The ordered pair is (-8, -6 )

→ Substitute x by -8

∴ y = $\frac{3}{4}$ (-8)

∴ y = -6

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (-8, -6) represents a point on the line

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1 year ago

You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

$t=\frac{95}{18}$

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

$j(t)=95t-9t^{2}$

$j'(t)=95-18t$

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

5 0

1 year ago