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AveGali [126]
3 years ago
7

I will give brainliest!!!

Mathematics
2 answers:
expeople1 [14]3 years ago
5 0

Answer:

<h3>CHOICE A.</h3>

Step-by-step explanation:

<em><u>There are 2 p’s and 3 q’s. So it is choice a.</u></em>

SOVA2 [1]3 years ago
3 0

Answer:

(choice A) 2p + 3q

Step-by-step explanation:

<em>hey there,</em>

<em />

< First, we should re-arrange q+p+q+p+q so that all the p's are together, and the q's are together so that it will be easier to solve (this step is obviously optional).

This is what you will get:

(q+q+q) + (p+p)

The way you solve this: the coefficient in front of every variable (p or q) is always ONE (1). Soo think of it this way: (1q+1q+1q) + (1p+1p).

If you pretended that there were no variables and it was just (1 + 1 + 1) = 3 and (1 + 1) = 2.

Now just add on your variables. 3q + 2p >

<u>Hope this helped! Feel free to ask any more questions if you still need help.</u>

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If z=2.5, x=102 and x=100, what is s?
Bad White [126]

I believe the problem ask for s which is the standard deviation. We must recall that the formula for z statistic is stated as:

z = (x – x over bar) / s

Where,

z = z statistic = 2.5

x = sample value or sample score = 102

x over bar = the sample mean or sample average  = 100

s = standard deviation = unknown

Rewriting the equation in terms of s:

s = (x – x over bar) / z

Substituting the given values into the equation:

s = (102 – 100) / 2.5

s = 0.8  

Therefore the standard deviation s is 0.8

7 0
3 years ago
Y=-2x-8 rate of change
Shtirlitz [24]
The rate of change is -2
3 0
3 years ago
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Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. F(x) =
Troyanec [42]

Answer:

a) P (x <= 3 ) = 0.36

b) P ( 2.5 <= x <= 3  ) = 0.11

c) P (x > 3.5 ) = 1 - 0.49 = 0.51

d) x = 3.5355

e) f(x) = x / 12.5

f) E(X) = 3.3333

g) Var (X) = 13.8891  , s.d (X) = 3.7268

h) E[h(X)] = 2500

Step-by-step explanation:

Given:

The cdf is as follows:

                           F(x) = 0                  x < 0

                           F(x) = (x^2 / 25)     0 < x < 5

                           F(x) = 1                   x > 5

Find:

(a) Calculate P(X ≤ 3).

(b) Calculate P(2.5 ≤ X ≤ 3).

(c) Calculate P(X > 3.5).

(d) What is the median checkout duration ? [solve 0.5 = F()].

(e) Obtain the density function f(x). f(x) = F '(x) =

(f) Calculate E(X).

(g) Calculate V(X) and σx. V(X) = σx =

(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge E[h(X)].

Solution:

a) Evaluate the cdf given with the limits 0 < x < 3.

So, P (x <= 3 ) = (x^2 / 25) | 0 to 3

     P (x <= 3 ) = (3^2 / 25)  - 0

     P (x <= 3 ) = 0.36

b) Evaluate the cdf given with the limits 2.5 < x < 3.

So, P ( 2.5 <= x <= 3 ) = (x^2 / 25) | 2.5 to 3

     P ( 2.5 <= x <= 3  ) = (3^2 / 25)  - (2.5^2 / 25)

     P ( 2.5 <= x <= 3  ) = 0.36 - 0.25 = 0.11

c) Evaluate the cdf given with the limits x > 3.5

So, P (x > 3.5 ) = 1 - P (x <= 3.5 )

     P (x > 3.5 ) = 1 - (3.5^2 / 25)  - 0

     P (x > 3.5 ) = 1 - 0.49 = 0.51

d) The median checkout for the duration that is 50% of the probability:

So, P( x < a ) = 0.5

      (x^2 / 25) = 0.5

       x^2 = 12.5

      x = 3.5355

e) The probability density function can be evaluated by taking the derivative of the cdf as follows:

       pdf f(x) = d(F(x)) / dx = x / 12.5

f) The expected value of X can be evaluated by the following formula from limits - ∞ to +∞:

         E(X) = integral ( x . f(x)).dx          limits: - ∞ to +∞

         E(X) = integral ( x^2 / 12.5)    

         E(X) = x^3 / 37.5                    limits: 0 to 5

         E(X) = 5^3 / 37.5 = 3.3333

g) The variance of X can be evaluated by the following formula from limits - ∞ to +∞:

         Var(X) = integral ( x^2 . f(x)).dx - (E(X))^2          limits: - ∞ to +∞

         Var(X) = integral ( x^3 / 12.5).dx - (E(X))^2    

         Var(X) = x^4 / 50 | - (3.3333)^2                         limits: 0 to 5

         Var(X) = 5^4 / 50 - (3.3333)^2 = 13.8891

         s.d(X) = sqrt (Var(X)) = sqrt (13.8891) = 3.7268

h) Find the expected charge E[h(X)] , where h(X) is given by:

          h(x) = (f(x))^2 = x^2 / 156.25

  The expected value of h(X) can be evaluated by the following formula from limits - ∞ to +∞:

         E(h(X))) = integral ( x . h(x) ).dx          limits: - ∞ to +∞

         E(h(X))) = integral ( x^3 / 156.25)    

         E(h(X))) = x^4 / 156.25                       limits: 0 to 25

         E(h(X))) = 25^4 / 156.25 = 2500

8 0
3 years ago
Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur
stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Three heads.
  • Event B: Fair coin.

The probability associated with 3 heads are:

  • 0.5^3 = 0.125 out of 0.5(fair coin).
  • 1 out of 0.5(biased).

Hence:

P(A) = 0.125 + 0.5 = 0.625

The probability of 3 heads and the fair coin is:

P(A \cap B) = 0.5(0.125) = 0.0625

Then, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

4 0
3 years ago
What is the definition of Net Worth?<br>​
Feliz [49]

Answer:

a

Step-by-step explanation:

5 0
3 years ago
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