First you need to divide 7 and 15, and the answer you get is 8, so 8+7=15. 8 will be your answer.
Answer:
Value of v that minimizes E is v = 3u/2
Step-by-step explanation:
We are given that;
E(v) = av³L/(v-u)
Now, using the quotient rule, we have;
dE/dv = [(v-u)•3av²L - av³L(1)]/(v - u)²
Expanding and equating to zero, we have;
[3av³L - 3av²uL - av³L]/(v - u)² = 0
This gives;
(2av³L - 3av²uL)/(v-u)² = 0
Multiply both sides by (v-u)² to give;
(2av³L - 3av²uL) = 0
Thus, 2av³L = 3av²uL
Like terms cancel to give;
2v = 3u
Thus, v = 3u/2
Draw a simple branch diagram to work the probabilities out.
You find that the chance of a poisonous mushroom is 0.08 and the chance of a red poisonous is 0.04.
So the probability that a poisonous mushroom is red is 1/2 or 0.5.
Answer:
Arla - $15, Kal - $17.5
Step-by-step explanation:
Let $x be Arla's hourly wage and $y be Kal's hourly wage.
1. In the first week on the job they each worked 40 hours and earned $(40x+40y) together. Their total earnings were $1300, so
40x+40y=1300.
2. In the second week on the job business was slow so Arla worked 20 hours and earned $20x, Kal worked 16 hours and earned $16y, $(20x+16y) in total. Their total pool earning was $580, so
20x+16y=580.
3. Solve the system of two equations:

Subtract them:

Then

Question: What's the joint frequency a 9th grader liked math?
<h3>Answer: 3%</h3>
Explanation:
Check out the attached image below. I've filled in the table with the missing values. The upper left corner is 3 because 7+3 = 10 in the first column. You'll fill in the other values in a similar fashion. Once we filled out the table, we can answer all of these questions. There are 3 ninth graders who like math out of 100 people total. So the final answer is 3/100 = 0.03 = 3%.
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Question: What's the marginal relative frequency of 9th graders surveyed?
<h3>Answer:
10%</h3>
Explanation:
Again, refer to the table to find that there are 10 ninth graders out of 100 total. So 10/100 = 0.10 = 10% is the answer.
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Question: Given a student was in 10th grade, what's the likelihood the student preferred English?
<h3>Answer:
46%</h3>
Explanation:
We only focus on the 10th graders because of the "given". There are 23 people in this row who like English out of 50 total sophomores. So 23/50 = 0.46 = 46% of the tenth graders liked English.
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Question: Given the subject was science, what's the likelihood the student was a freshman?
<h3>Answer:
72.5%</h3>
Explanation:
Focus on the science column only. There are 29 freshmen out of 40 students total. We get the percentage of 29/40 = 0.725 = 72.5%