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AnnyKZ [126]
3 years ago
6

Nick and Adam workout together. Nick weighs 160 pounds and is gaining about 3 pounds per week. Adam weighs 195 pounds and is los

ing about 2 pounds each week. Write an equation that can be used to find the number of weeks that it will take them to weigh the same amount.
Mathematics
1 answer:
Gekata [30.6K]3 years ago
4 0

Answer:

x=7

Step-by-step explanation:

(160+3x)=(195-2x)

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Segment AB has point A located at (6, 5). If the distance from A to B is 5 units, which of the following could be used to calcul
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The distance between (x1, y1) and (x2, y2) is found using the Pythagorean theorem. It is ...

\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For your values of (x1, y1) = (6, 5) and d = 5, substituting into the formula gives

\displaystyle 5=\sqrt{(x-6)^2+(y-5)^2}

This matches the first selection.

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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

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Question 2 289/50

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