HELP IM DESPERATE QUICK NO LINKS!!! DONT PUT LINKS

Can you guy's help me quick!

nadya68 [22]

Answer:

480,700 sets (first choice)

Step-by-step explanation:

If all 25 questions are different,

When order of selection counts:

P(25,7)

= 25!/(25-7)!

= 15511210043330985984000000/6402373705728000

= 2422728000

When order of selection does not count count:

C(25,7)

=25!/(7!*(25-7)!)

=15511210043330985984000000/(6402373705728000*5040)

= 480,700

Hope this helps, have a nice day.

6 0

3 years ago

What number is the opposite of the opposite of -99?

astraxan [27]

Answer:

Step-by-step explanation:

99

3 0

3 years ago

Read 2 more answers

Which description best describes the solution to the following system of equations?

matrenka [14]

Answer:the base root becomes 45

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Are the polygons similar? If they are, write a similarity statement and give the scale factor. The figures are not drawn to scal

Ugo [173]

The answer is C.
TUWV ~ DEFG; 6:4.5
please mark brainliest :)

7 0

3 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) Probability of distance at most 100m, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

Probability of distance at most 200:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

Probability of distance between 100 and 200:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

Distance exceeds the mean distance by more than 2σ:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

4 years ago