Answer :Anita's age = 7 years
Basilio's age = 2.5 years
Step 1. Assign variables to Anita's and Basilio's ages.
Anita's age = A
Basilio's age = B
Step 2. Use the information given to form two equations.
Anita is 4-1/2 years older than Basilio
A = 4.5 + B
Three times Anita's age added to six times Basilio's age is 36
3A + 6B = 36
Step 3. Substitute the value of A from first equation in to the second equation and solve for B
3(4.5 + B) + 6B = 36
13.5 + 3B + 6B = 36
13.5 + 9B = 36
9B = 36 - 13.5
9B = 22.5
B = 22.5/9 = 2.5
Step 4. Find the value of A
A = 4.5 + 2.5 = 7
Answer:
Anita's age = 7 years
Basilio's age = 2.5 years
Answer:
x = 60
Step-by-step explanation:
The sum of exterior angles in a rectangle is equal to 360
so to find the value of x we need to use all given values and write an equation:
x + 2x + x + 2x = 360 add like terms
6x = 360 divide both sides by 6
x = 60
<em>In decimal form, a = 114.5; b = 25.5</em>
We will write this as a system of equations, where 
and 
are the two angles.
We will use substitution to solve this system. We know what equals, so we plug that into the second equation.
Combine the like terms.
Add 
on both sides.
Divide both sides by 
.
Now just subtract that from 
.
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
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