Answer:
x=lindas age
3 times lindas age (3 time x or 3x) is decreased by 36 (-36), the result (=) is twice lindas age (2 times x or 2x)
now we have
3x-36=2x
subtract 2x
x-36=0
add 36
x=36
age=36
Step-by-step explanation:
The expected value of purchasing a raffle ticket is -0.647.
The expected value is derived in probability analysis by the multiplication of each of the potential possibilities by the likelihood that each result will occur and then adding all of those values together.
Using the formula below, we can calculate the expected value as follows:
However, the probability (Pr) of each event can be categorized as follows:
Thus;
Learn more about finding the expected value of a probability here:
Answer: The expected value of purchasing a raffle ticket would be $0.3449.
Step-by-step explanation:
Since we have given that
Number of tickets = 5500
Number of flower arrangements = 20
Number of gift certificates = 20
So, Probability that wining $70 at flower arrangement =
Probability that wining $25 at gift certificates =
So, Expected value of purchasing a raffle ticket would be
Hence, the expected value of purchasing a raffle ticket would be $0.3449.
Answer:
(1, -3)
(2, -5)
(3, -7)
Step-by-step explanation:
Randomly choose three values for "x". Substitute each of them separately into the formula to find the "y". Then write the solution as an ordered pair (x, y).
I will choose x=1, x=2 and x=3.
When x=1:
y = -2x - 1
y = -2(1) - 1
y = -2 - 1
y = -3
(1, -3)
When x=2:
y = -2x - 1
y = -2(2) - 1
y = -4 - 1
y = -5
(2 , -5)
When x=3:
y = -2x - 1
y = -2(3) - 1
y = -6 - 1
y = -7
(3, -7)
Three solutions for y = -2x - 1 are (0, -1), (2, -5), and (-3, 5), where x and y are related by this linear equation.
The equation y = -2x - 1 represents a linear relationship between x and y. To find three solutions for this equation, you can choose different values for x and calculate the corresponding values of y. Here are three solutions:
Solution 1:
Let x = 0:
y = -2(0) - 1
y = 0 - 1
y = -1
So, one solution is (0, -1).
Solution 2:
Let x = 2:
y = -2(2) - 1
y = -4 - 1
y = -5
So, another solution is (2, -5).
Solution 3:
Let x = -3:
y = -2(-3) - 1
y = 6 - 1
y = 5
So, a third solution is (-3, 5).
You can choose any values for x and calculate the corresponding y-values to find more solutions to this linear equation.
To learn more about linear equation here
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