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3 years ago

10

Can someone please help me?

2 answers:

krok68 [10]3 years ago

8 0

B, i think since it’s still showing 2/3 being multiples my n and then added to 12

ahrayia [7]3 years ago

7 0

Answer:

C

Step-by-step explanation:

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Convert. If necessary, round to the nearest tenth. (Recall: 1 in. = 2.54 cm) 16 cm . a. 4.5 c. 8.5 b. 6.3 d. 6.9

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3 years ago

Find f(-2) if f(x)= 2x^2-x+5

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What is the equation of a line, in point- slope form, that passes through (5, −3)(5, −3) and has a slope of 2323 ? y+5=23(x−3)y+

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Youyou havehave aa seriousserious doubledouble pastepaste problemproblem

the equation of the line that passes through the point (x1,y1) and has a slope of m is
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3 years ago

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Harry earns $45 a week more than Xena. George earns $18 a week less than Xena. If Harry earns the same amount in two weeks that

12345 [234]

Answer:

Xena earns $144 per a week.

Step-by-step explanation:

Harry earns - $ h

Xena earns - $ x

George earns -$ g

Harry earns $45 a week more than Xena. h= x+45

George earns $18 a week less than Xena. g = x - 18

If Harry earns the same amount in two weeks that George earns in three weeks

2h = 3g

So, we have the system of 3 equations:

h= x+45

g = x - 18

2h = 3g (*)

Substitute values of h and g into equation (*)

2(x+45) = 3(x-18)

2x + 90 = 3x - 54

90+54 = 3x- 2x

x= 144

5 0

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