Mike wants to rent a car for his vacation. The rental costs $125 a week plus $.15 a mile. How far can mike travel in a week if h
2 answers:
Answer: Mike can travel 500 miles if he wants to spend $200.
Step-by-step explanation: Given that Mike wants to rent a car for his vacation and the rental costs $125 a week plus $.15 a mile.
We are to find the number of miles that Mike can travel in a week if he wants to spend at the most $200.
Let x denotes the number of miles that Mike can travel.
Then, according to the given information, we have
Thus, Mike can travel 500 miles if he wants to spend $200.
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Answer:
Step-by-step explanation:
Let n be the number of tickets Makaya has to buy.
If the cost of one ticket is $1.5, then the cost of n tickets is $1.5n.
Makayla has $8 to buy tickets at the school fair, thus
The maximum number of tickets Makaya can buy is 5, so
Answer:
<em>(–81, 9), (–36, 6), (–1, 1) </em> are the correct three points.
Step-by-step explanation:
Given the function:
Please refer to the attached image.
The green line shows the graph of actual function.
It is reflected over y axis.
The reflected graph is shown in black color in attached image.
When reflected over y axis, the sign of variable changes from Positive to Negative.
So, the resultant function becomes:
i.e. we will have to give the values of x as negative now.
so, the options in which value of x is negative are the possible answers only.
The possible answers are:
(–81, 9), (–36, 6), (–1, 1) and
(–49, 7), (–18, 9), (–1, 1)
Now, we will check the square root function condition.
In the 2nd option, (–18, 9) does not satisfy the condition.
So, the correct answer is:
<em>(–81, 9), (–36, 6), (–1, 1)</em>
Answer:
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters
Step-by-step explanation:
<u>Step 1</u>:-
Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.
by Poisson random variable given that λ = 1.5 flaws/millimeter
Poisson distribution
<u>Step 2:</u>-
The probability that no flaws occur in a certain portion of wire
On simplification we get
P(x=0) = 0.223 flaws occur / millimeters
<u>Conclusion</u>:-
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters