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Ann [662]
3 years ago
10

James works as a tutor and a lawn mower. Both jobs require that he work at least three hours per week. Since James attends schoo

l, he can only work 15 hours a week. If James is paid $20 per hour as a tutor and $10 per hour as a lawn mower, which of the following is true about the number of hours he can work as a tutor (t) and lawn mower (m)?

Mathematics
1 answer:
charle [14.2K]3 years ago
4 0
Answer: The correct choice is that he can maximize his earnings by working 12 hours as a tutor and 3 hours as a lawn mower.

James will make the most money by working at the job that pays him the most. However, he has to work at least 3 hours at $10 per hour. 

Therefore, he should work only 3 mowing lawns and the rest tutoring.
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Describe the solution of g(x) shown in the graph.
Ugo [173]

Answer:

All real solutions

Step-by-step explanation:

  • The given graph is a maximum quadratic function.
  • The solution to the graph is where the graph intersects the x-axis.
  • We can see from the graph that, the function intersected the x-axis at two different points, hence its discriminant is greater than zero.
  • Hence the solution of g(x) are two distinct real solutions.
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3 years ago
The ratio of the numerator to the denominator of a fraction is 2 to 3. If both the numerator and the denominator are increased b
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(n + 2)/(d + 2) = 3/4  ------> 4n + 8 = 3d + 6
6 0
4 years ago
Read 2 more answers
-1 . (5) . (10) . (-4)
Ivanshal [37]
Sorry is this some sort of equation I’m confused
7 0
3 years ago
Here are summary statistics for randomly selected weights of newborn​ girls: nequals202​, x overbarequals28.3 ​hg, sequals6.1 hg
Lorico [155]

Answer:

The confidence interval is 27.5 hg less than mu less than 29.1 hg

(A) Yes, because the confidence interval limits are not similar.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17

7 0
4 years ago
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