Draw the picture and label the
width = w
The length of the monitor is six times the quantity of five less than half its width:
length = 6(w/2-5)
length = 3w-30
Area = (length)*(width)
384=(3w-30)*(w)
384=3w^2-30w
Answer:
the dimensions of the monitor in terms of its width is:
384=3w^2-30w
Answer:
$93.75
Step-by-step explanation:
The computation of the total cost of order is shown below:
The Cost of 8 pounds of Pepperoni is
= 8 × $3.15
= 25.2
The Cost of 15 pounds of cheese is
= 15 × $1.85
= $27.75
The Cost of 12 pounds of sausage is
= 12 × $3.40
= $40.8
So, The total cost is
= $25.2 + 27.75 + $40.80
= $93.75
Answer:
<em>The maximum number of kilowatt-hours is 235</em>
Step-by-step explanation:
<u>Inequalities</u>
Robert's monthly utility budget is represented by the inequality:
0.1116x + 23.77 < 50
Where x is the number of kilowatts of electricity used.
We are required to find the maximum number of kilowatts-hours used without going over the monthly budget. Solve the above inequality:
0.1116x + 23.77 < 50
Subtracting 23.77:
0.1116x < 50 - 23.77
0.1116x < 26.23
Dividing by 0.1116:
x < 26.23/0.1116
x < 235
The maximum number of kilowatt-hours is 235
Answer:
The probability that all three have type B+ blood is 0.001728
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they have type B+ blood, or they do not. The probability of a person having type B+ blood is independent of any other person. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability that a person in the United States has type B+ blood is 12%.
This means that 
Three unrelated people in the United States are selected at random.
This means that 
Find the probability that all three have type B+ blood.
This is P(X = 3).
The probability that all three have type B+ blood is 0.001728
