11x9=99
11x7=77
7x5/2=35/2
99+99+77+35/2+35/2= 310
11x7=77
7x5/2=35/2
99+99+77+35/2+35/2= 310
Which is the graph of y = 2.5x? A. B. C. D.
1 answer:
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Answer:
Amounts in the intervall
Step-by-step explanation:
We will take the variables c and s, where
c = amount to spend that month
s = savings of the month.
Since each month we save at least 100, we know that
Moreover, we earn $250 per month and we spend $20 in a kayaking club, which tells us that
and so
By multiplying by -1, the first inequality also tells us that
Consequently by adding 230 to the inequality, we get that
You put the numbers you want to compare in a line with a space in the middle. the put
if the left is bigger and
if the right side is bigger.
if the left is bigger and
if the right side is bigger.
Answer:
Step-by-step explanation:
Hello!
The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.
Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.
Be:
X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)
n₁= 12
X[bar]₁= 20.50 hs
S₁²= 199.76hs² (S₁= 14.13hs)
X₂: Battery duration between charges when the primary usage of the phone is internet applications.
n₂= 10
X[bar]₂= 8.50
S₂²= 33.29hs² (S₂= 5.77hs)
Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)
The parameters of interest are σ₁² and σ₂²
a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1
The hypotheses are:
H₀: σ₁²/σ₂² ≤ 1
H₁: σ₁²/σ₂² > 1
There is no level of significance listed so I've chosen α: 0.05
b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:
S²= *[∑X²-(∑X)²/n]
For the first sample:
n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36
S₁²= *[7240.36-(246)²/12]= 199.76hs²
S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs
For the second sample:
n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12
S₂²= *[1022.12-(85)²/10]= 33.2911hs²
S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs
c)
For this hypothesis test, the statistic to use is a Snedecors F:
This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:
The rejection region is then F ≥ 3.10
p-value: 0.006
Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.
Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.
I hope this helps!
