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Paladinen [302]
3 years ago
13

14

Mathematics
1 answer:
notka56 [123]3 years ago
8 0

Answer:

yes razi is correct pateran 3

Step-by-step explanation:

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Type the verb form suggested by the related word(s).<br><br> exerciser
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Answer:

exercising

Step-by-step explanation:

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3 years ago
Solve the equation using the quadratic formula
amm1812

Answer:

Step-by-step explanation:

Rewrite this quadratic equation in standard form:  2n^2 + 3n + 54 = 0.  Identify the coefficients of the n terms:  they are 2, 3, 54.

Find the discriminant b^2 - 4ac:  It is 3^2 - 4(2)(54), or -423.  The negative sign tells us that this quadratic has two unequal, complex roots, which are:

     -(3) ± i√423        -3 ± i√423

n = ------------------- = ------------------

             2(2)                      4

3 0
2 years ago
Please help me with part b of this question. THanks.
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3 years ago
Expressions is equal to 2(40 - p)?
Bumek [7]
80 - 2p
You use the distributive property
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1 year ago
(1 point) An insurance company offers its policyholders a number of different premium payment options. For a randomly selected p
Olegator [25]

Answer:

a)

X    |  1        3      5       7

f(X) | 0.4   0.2   0.2    0.2

b) P(4

Step-by-step explanation:

For this case we have defined the cumulative distribution function like this:

F(X) = 0, x

F(X) = 0.4, 1 \leq x

F(X) = 0.6, 3 \leq x

F(X) = 0.8, 5 \leq x

F(X) = 1, x \geq 7

And we know that the general definition for the distribution function is given by:

F(x) = P(X \leq x) = \sum_{i\leq k} f(i)

Where f represent the density function.

Part a

For this case we need to find the density function, so we can find the values for the density for each value of X = 1,2,3,4,5,6,7,... since X is a discrete random variable.

f(1) = P(X=1) = P(X \leq 1) - P(X=0) = F(1) -F(0) = 0.4-0=0.4

f(2) = P(X=2) = P(X \leq 2) - P(X=0)- P(X=1) = F(2) -F(1) = 0.4-0.4=0

f(3) = P(X=3) = P(X \leq 3) - P(X=0)- P(X=1) -P(X=2) = F(3) -F(2) = 0.6-0.4=0.2

f(4) = P(X=4) = P(X \leq 4) - P(X=0)- P(X=1) -P(X=2)-P(X=3) = F(4) -F(3) = 0.6-0.6=0

f(5) = P(X=5) = P(X \leq 5) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4) = F(5) -F(4) = 0.8-0.6=0.2

f(6) = P(X=6) = P(X \leq 6) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5) = F(6) -F(5) = 0.8-0.8=0

f(7) = P(X=7) = P(X \leq 7) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5)-P(X=6) = F(7) -F(6) = 1-0.8=0.2

And for any value higher than 7 we have that:

x_i \in [8,9,10,...]

f(x_i) = F(X_i) -F(X_i -1) = 1-1=0

So then we have our density function defined like this:

X    |  1        3      5       7

f(X) | 0.4   0.2   0.2    0.2

Part b

For this case we want to find this probability P(4

And since the random variable is discrete we can write this like that:

P(4

5 0
3 years ago
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