Answer: 200 minutes have to be used for the costs of both plans to be the same.
Step-by-step explanation:
Let x represent the number of minutes that have to be used for the costs of both plans to be the same.
Package A is $35.00 per month with an additional charge of $0.15 per minute for long distance. This means that the cost of using package A for x minutes in a month would be
35 + 0.15x
Package B is $45.00 per month with an additional charge of $0.10 per minute for long distance. This means that the cost of using package A for x minutes in a month would be
45 + 0.1x
For both costs to be the same, it means that
35 + 0.15x = 45 + 0.1x
0.15x - 0.1x = 45 - 35
0.05x = 10
x = 10/0.05
x = 200
Answer:
If we look at ΔBDH, we notice that m∠BHG has to be 108 - 36 - 39 = 105°. Because x and ∠BHG are a linear pair we know that x = 180 - ∠BHG = 180 - 105 = 75°.
Your answer would be 3160.
According to the attached formula:
apothem = side length / 2 * tan (180/number of sides)
apothem = 6 / 2 * tan (180/10)
apothem = 6 / 2 * 0.32492
apothem =
<span>
<span>
<span>
9.23 cm
</span></span></span>
<span><span /></span>
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 17
For the alternative hypothesis,
µ < 17
This is a left tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 80,
Degrees of freedom, df = n - 1 = 80 - 1 = 79
t = (x - µ)/(s/√n)
Where
x = sample mean = 15.6
µ = population mean = 17
s = samples standard deviation = 4.5
t = (15.6 - 17)/(4.5/√80) = - 2.78
We would determine the p value using the t test calculator. It becomes
p = 0.0034
Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.
The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
