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umka21 [38]
3 years ago
8

Use the distributive property to write the expression without parentheses -(5y + 6)

Mathematics
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

-5y -6

Step-by-step explanation:

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Of the specific numbers of each color, red has the most marbles, so mathematically speaking, red has the highest chance of being chosen first

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What is the answers for question 5 and 7?????? Having a bit of trouble and I’m in need of help! Please and thank you!
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3 years ago
Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.
Sindrei [870]

Answer:

a) P(X=3) = 0.1

b) P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

c) P(X=4) = 0.3

d) P(X=0) = 0.2

e) E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

f) E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

Step-by-step explanation:

We have the following distribution

x      0     1     2   3   4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

P(X=3) = 0.1

Part b

We want this probability:

P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

Part c

For this case we want this probability:

P(X=4) = 0.3

Part d

P(X=0) = 0.2

Part e

We can find the mean with this formula:

E(X)= \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

Part f

We can find the second moment with this formula

E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

4 0
4 years ago
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