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dusya [7]
3 years ago
12

4

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
7 0

Answer:

b

Step-by-step explanation:

because

You might be interested in
The midpoint of is (14, 26). One endpoint of is A(32, 40). What are the coordinates of B?
cupoosta [38]
For x,
(32+x)/2=14
32+x=14*2
32+x=28
x=28-32
x=-4

For y,
(40+y)/2=26
40+y=26*2
40+y=52
y=52-40
y=12

Therefore coordinates of B(-4,12)
Hope this helps!
7 0
2 years ago
PLEASE HELP!! WILL MARK BRAINLIEST TO WHOEVER GET IT RIGHT
Vinvika [58]

Answer:

neither, since the gradients are not the same, as well as the c value

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
If y varies directly as x and y=32 when x=4 find the value of y when x=5
Alex Ar [27]

Answer:

y=40

Step-by-step explanation:

32 divided by 4 equals 8. And 5 times 8 equals 40.

3 0
2 years ago
We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent
Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(c) P(X = −2) = 1/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

(c) P(Y = 2) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

                  ( X = -2 ):  { X_2 = 1 , X_1 = 3 }

                  P ( X = -2 ):  P ( X_2 = 1 ) * P ( X_1 = 3 )

                                 :  ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 1 / 9 )

   

                  ( X = -1 ):  { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

                 P ( X = -1 ):  P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 2 / 9 )

         

       ( X = 0 ):  { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } +  { X_2 = 3 , X_1 = 3 }

       P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 3 / 9 ) = ( 1 / 3 )

       

                    ( X = 1 ):  { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

                 P ( X = 1 ):  P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 2 / 9 )

                    ( X = 2 ):  { X_2 = 1 , X_1 = 3 }

                  P ( X = 2 ):  P ( X_2 = 3 ) * P ( X_1 = 1 )

                                    :  ( 1 / 3 ) * ( 1 / 3 )

                                    : ( 1 / 9 )                  

- The distribution Y = X_2,

                          P(Y=0) = 0

                          P(Y=1) =  1/3

                          P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

7 0
3 years ago
Helen plays basketball. For free throws, she makes the shot 78% of the time. Helen must now attempt two free throws. C = the eve
riadik2000 [5.3K]

Answer:

0.6708 or 67.08%

Step-by-step explanation:

Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:

P(C+D) =  P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708

There is a 0.6708 probability that Helen makes both free throws.

5 0
3 years ago
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