Using the given functions, it is found that:
- Lower total cost at Jump-n-Play: 40, 64.
- Lower total cost at Bounce house: 28, 8, 30.
- Same total cost at both locations: 32.
<h3>What are the cost functions?</h3>
For n visits to Jump-n-play, the cost is:
J(n) = 189 + 3n.
For n visits to Bounce Word, the cost is:
B(n) = 125 + 5n.
Comparing them, we have that:
Hence:
- For less than 32 visits, the cost at Bounce World is lower.
- For more than 32 visits, the cost at Jump-n-play is lower.
Hence:
- Lower total cost at Jump-n-Play: 40, 64.
- Lower total cost at Bounce house: 28, 8, 30.
- Same total cost at both locations: 32.
More can be learned about functions at brainly.com/question/25537936
Answer:
y=-2x-12
Step-by-step explanation:
first, putt 3x-6y=2 in standard form:
subtract 3x from both sides: -6y=-3x+2
divide both sides by -6 to isolate y: y=1/2x-1/3
if the other line is perpendicular to this, you must find the slope by finding the opposite reciprocal of 1/2x: -2
now we have 2 points a slope, so we use the point-slope formula: y-y1=m(x-x1)
y+2=-2(x+5)
Use the distributive property: y+2=-2x-10
subtract 2 for both sides to isolate y: y=-2x-12
Answer:
z = 66
Step-by-step explanation:
QT is a midsegment. Thus, applying the midsegment theorem:
RS = 2(QT)
QT = z - 33,
RS = z
Plug in the values into the equation
z = 2(z - 33)
z = 2z - 66
Subtract 2z from each side
z - 2z = -66
-z = -66
Divide both sides by -1
z = 66
Answer:
Claim : men weigh of wild jackalopes is 69.9
The null hypothesis : H0 : μ = 69.9
Alternative hypothesis : H1 : μ ≠ 69.9
Test statistic = −2.447085
P value = 0.0174
Conclusion :
Fail to Reject the Null hypothesis
Step-by-step explanation:
From the question given :
The claim is that : mean weight of wild jackalopes is still the same as 10 years with a mean weight of 69.9 lbs.
The null hypothesis : H0 : μ = 69.9
Alternative hypothesis : H1 : μ ≠ 69.9
Using calculator :
Sample mean (x) = 66
Sample standard deviation (s) = 12.345
The test statistic t :
(x - μ) / (s/√n)
n = sample size = 60
(66 - 69.9) / (12.345 / √60)
t = −2.447085
P value at α 0.01, df = 59 is 0.0174
Since the p value is > 0.01, the result is not significant at 0.01. Therefore, we fail to reject the Null
