The first thing that we have to do is to assign variables to the cat's and the dog's ages. Let us use c to represent the age of the cat and d for the dog's age. So, let's translate the word problem into a mathematical one.
c = d + 3
This is the first equation while the second equation is
(d+3)*d = 54
which becomes
d^2 +3d -54 = 0
which can be factored into (d-6)(d+9). This gives us two roots which are 6 and -9. However, age is a non-negative number so we will take 6 as the age of Helen's dog. Thus, making Sally's cat 9 years old.
Answer:
58/33
Step-by-step explanation:
If Pipe 1 (P1) takes x hours to fill the pool, Pipe 2 (P1) and pipe 2 (P2) takes (x-9) hours to fill the pool, and pipe 2 (P2) takes (x+7) hours to fill the pool.
That is,
P1 = x hrs
P1+P2 = (x-9) hrs
P3 = (x+7) hrs
In 1 hour, P1 fills 1/x of the pool, P1+P2 fills 1/(x-9) of the pool and P2 fills 1/(1+7) of the pool.
Therefore,
1/x+1/(1+7) = 1/(x-9) => ((x+7)+x)/(x)(x+7)=1/(x-9) => (2x+7)/x^2+7x = 1/(x-9) => (2x+7)(x-9)=x^2+7x => x^2-18x-63 =0
Solving for x
x= (-b+/- sqrt (b^2-4ac)/2a, where a=1, b=18, and c=63
Substituting;
x1=21 and x2=-3 (the negative x is ignored as it does not make sense).
Therefore, x = 21
This means,
P1 takes 21 hours to fill the pool
P1+P2 takes (21-9) hours = 12 hours to fill the pool while P3 takes (21+7) hours = 28 hours
Answer:
A confidence interval for a mean should be constructed to estimate the variable of interest.
Step-by-step explanation:
The confidence interval for a mean uses the sample mean to estimate the true population mean. Since it is a confidence interval, instead of a single number for the mean, the result shows a lower estimate and an upper estimate of the mean. In this experiment, the confidence interval cannot be for a proportion since the population of healthy males is unknown, but the calculation is being done with a sample of 75 healthy males.
