Hey there! I'm happy to help!
Let's have the adult tickets be a and the children tickets be c.
a=2c (twice as many adult tickets as child tickets)
5a+2c=700 (dollar value of each ticket)
Have a wonderful day! I hope that the helps! :D
This can be solved by factoring.
First, set the expression equal to zero.

Then, find two the factors of

whose sum is

.

Split

into these two factors.

Next, factor by grouping.

By the Zero Product Property, set each factor equal to zero.


These are the solutions. The Complex Conjugate Root Theorem and the Fundamental Theorem of Algebra both state that, in essence, real and imaginary solutions come in pairs of two and every polynomial of degree

has exactly

complex roots, but real roots are also complex roots. That sounds confusing, but this just means that you're done.
Your answers are -2 and 1/3. There are two real roots.
Answer:
<u>The Answer in Exact Form:</u>
-3/22
<u>In Decimal Form:</u>
-0.136
Step-by-step explanation:
-
÷5
=-3/22
Answer:
$9,220,000(0.888)^t
Step-by-step explanation:
Model this using the following formula:
Value = (Present Value)*(1 - rate of decay)^(number of years)
Here, Value after t years = $9,220,000(1 -0.112)^t
Value after t years = $9,220,000(0.888)^t
Answer:
Probability that average height would be shorter than 63 inches = 0.30854 .
Step-by-step explanation:
We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.
Also, a random sample of 4 women from this population is taken and their average height is computed.
Let X bar = Average height
The z score probability distribution for average height is given by;
Z =
~ N(0,1)
where,
= population mean = 64 inches
= standard deviation = 4 inches
n = sample of women = 4
So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)
P(X bar < 63) = P(
<
) = P(Z < -0.5) = 1 - P(Z <= 0.5)
= 1 - 0.69146 = 0.30854
Hence, it is 30.85% likely that average height would be shorter than 63 inches.