The sum of liz's and melinda's ages is 28. if liz was 1/3 melinda's age 6 years ago, how old will liz be next year?
1 answer:
The first thing we must do for this case is to define variables:
We have then:
x: liz's age
y: melinda's age
We now write the system of equations:
By solving the system graphically we have that the age of each one is currently:
Next year, the age of liz is:
Answer:
Next year, Liz's age is 11 years.
Note: see attached image for the graphic solution.
We have then:
x: liz's age
y: melinda's age
We now write the system of equations:
By solving the system graphically we have that the age of each one is currently:
Next year, the age of liz is:
Answer:
Next year, Liz's age is 11 years.
Note: see attached image for the graphic solution.
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Answer:
a) 0.69
The probability that a randomly selected 10-year old child will be more than 51.75 inches tall
P(X>51.75 ) = 0.6915
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
<em>Given mean of the Population = 54.6 inches</em>
<em>Given standard deviation of the Population = 5.7 inches</em>
<em>Let 'X' be the random variable of normal distribution</em>
Let 'X' = 51.75 inches
<u><em>Step(ii):</em></u>-
<em>The probability that a randomly selected 10-year old child will be more than 51.75 inches tall</em>
<em>P(X>51.75 ) = P(Z>-0.5)</em>
= 1 - P( Z < -0.5)
= 1 - (0.5 - A(-0.5))
= 1 -0.5 + A(-0.5)
= 0.5 + A(0.5) (∵A(-0.5)= A(0.5)
= 0.5 +0.1915
= 0.6915
<u><em>Conclusion</em></u>:-
<em>The probability that a randomly selected 10-year old child will be more than 51.75 inches tall</em>
<em>P(X>51.75 ) = 0.6915</em>