How do you get (-6,6), (9,1) in Point-Slope Form?

2. Simplify: (-4a+b-2c) - (3a +26-c) -)

erastovalidia [21]

Answer:

I thinks its, -7a+b-3c+26

Step-by-step explanation:

idk

5 0

2 years ago

7g - 6 = -20 what is the answer

Brut [27]

7g - 6 = -20

Add 6 to both sides

7g = -14

Divide both sides by 7
g = -2

answer: g = -2

7 0

1 year ago

Read 2 more answers

Diego’s dog weighs more than 10 kilograms and less than 15 kilograms. Select the set of inequalities that must be true if w is t

hichkok12 [17]

Answer:

Step-by-step explanation:

Correct

w > 10

Incorrect

w < 10

Incorrect

w > 11

Incorrect

w < 11

Correct

w < 15

w < 15Correct

w > 10

Incorrect

w < 10

Incorrect

w > 11

Incorrect

w < 11

Correct

w < 15

w < 15Correct

w > 10

Incorrect

w < 10

Incorrect

w > 11

Incorrect

w < 11

Correct

w < 15

w < 15Correct

w > 10

Incorrect

w < 10

Incorrect

w > 11

Incorrect

w < 11

Correct

w < 15

w < 15Correct

w > 10

Incorrect

w < 10

Incorrect

w > 11

Incorrect

w < 11

Correct

w < 15

w < 15Correct

w > 10

Incorrect

w < 10

Incorrect

w > 11

7 0

2 years ago

For which inequality is {x | x R, x > 6} the solution set?

CaHeK987 [17]

Answer:

C

Step-by-step explanation:

B: 2X + 12 > 3

2x > -9

x> -9/2

C: 2X - 5 > 7

2x > 12

x > 6

D: 2x - 6 > 24

2x > 30

x > 15

Hope this helps ^-^

6 0

2 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

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 time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

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

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago