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ArbitrLikvidat [17]
3 years ago
6

Question 20 points if correct

Mathematics
2 answers:
True [87]3 years ago
7 0
Is This On Khan Academy because I be needing help
Vlada [557]3 years ago
5 0

Answer:

B = 155° , C = 25° , D = 155°

Step-by-step explanation:

Given

A = 25°

C = A = 25° [ Vertically Opposite angles ]

Now

A + B = 180°( being linear pair)

25° + B = 180°

B = 180° - 25°

B = 155°

Also

D = B = 155°( vertically opposite angles )

You might be interested in
Wayne Gretsky scored a Poisson mean six number of points per game. sixty percent of these were goals and forty percent were assi
BartSMP [9]

Answer:

a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

b) 0.05 = 5% probability that he has four goals and two assists in one game

Step-by-step explanation:

In hockey, a point is counted for each goal or assist of the player.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval. The standard deviation is the square root of the mean.

(a) Find the mean and standard deviation for the total revenue he earns per game.

60% of six are goals, which means that 60% of the time he earned 3K.

40% of six are goals, which means that 40% of the time he earned 1K.

The mean is:

\mu = 6*0.6*3 + 6*0.4*1 = 13.2

The standard deviation is:

\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63

The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

(b) What is the probability that he has four goals and two assists in one game

Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.

Probability of four goals:

60% of 6 are goals, which means that:

\mu = 6*0.6 = 3.6

The probability of scoring four goals is:

P(A) = P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.19122

Probability of two assists:

40% of 2 are assists, which means that:

\mu = 6*0.4 = 2.4

The probability of getting two assists is:

P(B) = P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.26127

Probability of four goals and two assists:

P(A \cap B) = P(A)*P(B) = 0.19122*0.26127 = 0.05

0.05 = 5% probability that he has four goals and two assists in one game

5 0
2 years ago
Simplify: 5x + 4 + 9x​
Klio2033 [76]
The correct answer is; 14x +4
8 0
3 years ago
Read 2 more answers
Jarvis invested some money at 6% interest. Jarvis also invested $58 more than 3 times that amount at 9%. How much is invested at
DerKrebs [107]

9514 1404 393

Answer:

  • $3309 at 6%
  • $9985 at 9%

Step-by-step explanation:

Let x and y represent amounts invested at 6% and 9%, respectively.

  y = 3x +58 . . . . . . . the amount invested at 9%

  0.06x +0.09y = 1097.19 . . . . . . total interest earned

__

Substituting for y, we have ...

  0.06x +0.09(3x +58) = 1097.19

  0.33x + 5.22 = 1097.19 . . . . . . . . . simplify

  0.33x = 1091.97 . . . . . . . . . . . . subtract 5.22

  x = 3309 . . . . . . . . . . . . . . . . divide by 0.33

  y = 3(3309) +58 = 9985

$3309 is invested at 6%; $9985 is invested at 9%.

8 0
3 years ago
ASAP!!!!!!!!!!!!!!!!!!!
steposvetlana [31]

Answer:

V = 2143.57 cm^3

Step-by-step explanation:

We want to find the volume of the sphere

V = 4/3 pi r^3 where r is the radius and pi = 3.14

V = 4/3 ( 3.14) ( 8)^3

V = 2143.57333 cm^3

Rounding to the nearest hundredth

V = 2143.57 cm^3

3 0
3 years ago
Read 2 more answers
Two fair dice, one blue and one red, are tossed, and the up face on each die is recorded. Define the following events:
ZanzabumX [31]

Answer:

Step-by-step explanation:

Given that two fair dice, one blue and one red, are tossed, and the up face on each die is recorded.

a) P(E) = P(the difference of the numbers is 3 or more}

Favourable events are (1,4) (1,5)(1,6) (2,5) (2,6) (3,6) (4,1) (5,1) (5,2) (6,1) (6,2)(6,3)

P(E) = \frac{12}{36} =\frac{1}{3}

b)P(F)

Favourable events for F = (1,1) (2,2)...(6,6)

P(F) = \frac{6}{36} =\frac{1}{6}

c) P(EF)

There is no common element between E and F

P(EF) =0

4 0
3 years ago
Read 2 more answers
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