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3 years ago

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A basketball player is fouled while attempting to make a basket and receives two free throws. The opposing coach believes there

is a 30% chance that the player will miss both shots, a 16% chance that he will make one of the shots, and a 54% chance that he will make both shots.A. Construct the appropriate probability distribution. B. What is the probability that he makes no more than one of the shots?C. What is the probability that he makes at least one of the shots?

1 answer:

vladimir1956 [14]3 years ago

7 0

Answer: i cant draw A here but B is 84%

Step-by-step explanation:

30%+54%=84% (i think)

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Draw a conclusion for the statement if possible. If it is Monday, then the library is closed. Today is Tuesday

USPshnik [31]

Answer:

D. It is not possible to draw a conclusion

Step-by-step explanation:

we are not given enough information about the library schedule.

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3 years ago

Find the lowest common denominator 3x^3y^2 and 18xy^3

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3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

Which pair of expressions represents inverse functions

Natali5045456 [20]

Answer: C. $\dfrac{x+3}{4x-2}$ and $\dfrac{2x+3}{4x-1}$

Step-by-step explanation: if a function f(x) has g(x) as its inverse then it satisfies fog(x)=x and gof(x)=x

C. f(x)= $\dfrac{x+3}{4x-2}$ and g(x)= $\dfrac{2x+3}{4x-1}$

fog(x)=f( $\dfrac{2x+3}{4x-1}$ )

= $\dfrac{\dfrac{2x+3}{4x-1}+3 }{\dfrac{4(2x+3)}{4x-1}-2 }$

=x

gof(x)=g( $\dfrac{x+3}{4x-2}$ )

= $\dfrac{\dfrac{2(x+3)}{4x-2} +3}{\dfrac{4(x+3)}{4x-2}-1 }$

=x

hence C. is the pair of inverse functions

7 0

3 years ago

Read 2 more answers

Expanded -8(-2x-5)<br> please give answer

nika2105 [10]

Answer:

-80

Step-by-step explanation:

We have to do the parentheses first because it's part of PEDMAS.

-2x-5= 10

Next, we take 10 times -8

-8 x 10= -80

There you go! Hope this helps!

8 0

3 years ago

Read 2 more answers