If barbara used the washer for 3 hours and her brother returned it to the store 7 hours after she rented it how much does her br
other owe her
2 answers:
Answer:
a. $15.00
Step-by-step explanation:
okay so the complete question is
After a heavy rainstorm, Barbara’s dog left muddy paw prints all over her back patio. She decided to rent an electric pressure washer to clean up the mess. The piece wise function below shows the rental charges for the pressure washer, dependent upon how many hours she rents it. Due to high demand, you cannot rent the washer for more than 24 hours at a time. Barbara’s brother asked to use the washer and agreed to pay Barbara for any additional monies owed.
(See attached picture)
If Barbara used the washer for 3 hours and her brother returned it to the store 7 hours after she rented it, how much does her brother owe her?
$15.00
$41.00
$60.00
$61.00
Since Barb and her bro used it for 7 hours total, we'll be using the second function, or 26 = 5(h-4)
plug 7 in for h and solve and you'll get
41
subtract 26 (got that from the first function) from 41 and you'll get your answer:
$15.00
You might be interested in
Answer:
15.6%
Step-by-step explanation:
Since each day there is a 6% chance that Lisa smiles at him then that means that each day there is a 94% chance that Lisa does not smile at him. To find the probability of Milhouse going longer than a month (30 days) without a smile from Lisa we need to multiply this percentage in decimal form for every day of the month. This can be solved easily by putting 94% to the 30th power which would be the same, but first, we need to turn it into a decimal...
94% / 100 = 0.94
= 0.156
Now we can turn this decimal into a percentage by multiplying by 100
0.156 * 100 = 15.6%
Finally, we can see that the probability that Milhouse goes longer than a month without a smile from Lisa is 15.6%
Answer:
- When Two containers are selected
(a) Probability that the second one selected is defective given that the first one was defective = 0.00450
(b) Probability that both are defective = 0.0112461
(c) Probability that both are acceptable = 0.986
2. When Three containers are selected
(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.
(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451
(c) Probability that all three are defective = 6.855 x .
Step-by-step explanation:
We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.
Total containers = 445
Defective ones = 3
Non - Defective ones = 442 { Acceptable ones}
- Two containers are selected, at random, without replacement from the batch.
(a) Probability that the second one selected is defective given that the first one was defective is given by;
<em>Since we had selected one defective so for selecting second the available </em>
<em> containers are 444 and available defective ones are 2 because once </em>
<em> chosen they are not replaced.</em>
Hence, Probability that the second one selected is defective given that the first one was defective = = 0.00450
(b) Probability that both are defective = P(first being defective) +
P(Second being defective)
= = 0.0112461
(c) Probability that both are acceptable = P(First acceptable) + P(Second acceptable)
Since, total number of acceptable containers are 442 and total containers are 445.
So, Required Probability = = 0.986
- Three containers are selected, at random, without replacement from the batch.
(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;
<em>Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .</em>
Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = = 0.002.
(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;
<em>Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.</em>
Hence, Required probability = = 0.00451 .
(c) Probability that all three are defective = P(First being defective) +
P(Second being defective) + P(Third being defective)
= = 6.855 x
.