Part A: 5 is the coefficient, w is the variable, 50 is the constant

Part B: 10 x 5 = 50 + 2 x 20 = 40+ 50 = $140

Part C: the constant,50,would change and become 85. Hope this helps :)

8 0

3 years ago

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If a die is rolled one time, find probability of getting a number greater than 3 and an odd number.

Nadya [2.5K]

Answer:

$\dfrac{1}{6}$ .

Step-by-step explanation:

It is given that a die is rolled one time. So,

Sample space : $S=\{1,2,3,4,5,6\}$

Number is greater than 3 : $A=\{4,5,6\}$

Number is odd : $A=\{1,3,5\}$

Number greater than 3 and an odd number : $A\cap B=\{5\}$

It is conclude that,

$n(S)=6, n(A)=3,n(B)=3,n(A\cap B)=1$

The probability of getting a number greater than 3 and an odd number is

$P=\dfrac{n(A\cap B)}{n(S)}$

$P=\dfrac{1}{6}$

Therefore, the required probability is $\dfrac{1}{6}$ .

4 0

3 years ago

Find the limits of integration ly, uy, lx, ux, lz, uz (some of which will involve variables x,y,z) so that ∫uz lz∫uxlx∫uyly????y

SSSSS [86.1K]

Answer:

X from 0 to 21

Y from 0 to 7

Z from 0 to 3

Step-by-step explanation:

Since we are being asked by the integration limits in first octant (positive x, positive y and positive z) we need to know where does the plane intersect this axes. For this we have:

for x=0 and y=0

7z=21

z=3

for x=0 and z=0

3y=21

y=7

for z=0 and y=0

x=21

This means that the integration limits are:

X from 0 to 21

Y from 0 to 7

Z from 0 to 3

8 0

3 years ago

11. Suppose you buy 1 ticket for $1 out of a lottery of 1,000 tickets where the prize for the one winning ticket is to be $500.

erastovalidia [21]

Answer:

-50 cents

Step-by-step explanation:

The "Winnings" random variable has values -1 and 499

The associated probabilities are P(-1)= 999/1000; P(499)=1/1000

-----------------------------

Expected Value = -1(999/1000) + 499(1/1000) = -500/1000 = -50 cents

6 0

2 years ago

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