Please help!!! Will mark brainlist
1 answer:
Answer:
Not a Solution
Step-by-step explanation:
We are given two inequalities which are
y ≤ x -4 .............(i)
-x+3y>-4 ............(ii)
Also we are given an ordered pair which is (5, 1/3)
Now from this order pair we see that
x = 5 and y = 1/3
Because in an ordered pair the first element represents the x value while the second value represent the y value
Now to find whether this order pair satisfies the given inequality or not we have to plugin the values of x and y in both inequalities separately and see whether it satisfies the in equality or not
Taking First inequality:
which is
y ≤ x -4
Putting x = 5 and y = 1/3 in inequality
it becomes
≤ 5 -4
≤ 1 ∵ which is true
So this inequality holds the order pair
Taking second inequality:
which is
-x+3y> -4
Putting x = 5 and y = 1/3 in inequality
it becomes
-5+ > -4
-5+1 >-4
-4>-4 ∵ which is false because - 4 = - 4
So this inequality does not holds the order pair
So the order pair is not solution of the given inequalities because of the reason that second inequality is not satisfied
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Answer:
Step-by-step explanation:
To find the inverse of a function, what I like to do is to switch the places of x and y, and then try to isolate y again like so:
Hope this helps!
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!
