Select all input values for which h(x)=2

Write a decimal where the value of a digit is one tenth as much as a digit in another place

uysha [10]

Jabshudoa wvduoslabdu9sbwbd

4 0

3 years ago

I need help please

nlexa [21]

Answer:

49

Step-by-step explanation:

No she is not correct because we dont have only one angle angle G is 90.

So we can easily find the angle D

180-(90+41) =49

7 0

2 years ago

Can someone help with this please

nasty-shy [4]

Answer:

22x^2 -10xy + 12x

Step-by-step explanation:

To add and subtract polynomials, combine only like terms. When doing it vertically, stack polynomials and line up the same bases and exponents. Once this is done, simply add and subtract the coefficients.

10x^2 + 12xy + 4x

+ 12x^2 -22xy + 8x

_________________

22x^2 -10xy + 12x

Here we added 10 + 12 = 22, 12 + -22 = -10 and 4 + 8 = 12.

6 0

3 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

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.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

8 0

3 years ago

Read 2 more answers

Which equations represent circles that have a diameter of 12 units and a center that lies on the y-axis? Select two options

Sholpan [36]

Answer:

The answer is A and E

x2+(y−3)2=36

x^{2}+(y+8)^{2}=36x2+(y+8)2=36

4 0

3 years ago

Read 2 more answers