Answer:
(A) Yes, since the test statistic is in the rejection region defined by the critical value, reject the null. The claim is the alternative, so the claim is supported.
Step-by-step explanation:
Null hypothesis: The wait time before a call is answered by a service representative is 3.3 minutes.
Alternate hypothesis: The wait time before a call is answered by a service representative is less than 3.3 minutes.
Test statistic (t) = (sample mean - population mean) ÷ sd/√n
sample mean = 3.24 minutes
population mean = 3.3 minutes
sd = 0.4 minutes
n = 62
degree of freedom = n - 1 = 62 - 1 = 71
significance level = 0.08
t = (3.24 - 3.3) ÷ 0.4/√62 = -0.06 ÷ 005 = -1.2
The test is a one-tailed test. The critical value corresponding to 61 degrees of freedom and 0.08 significance level is 1.654
Conclusion:
Reject the null hypothesis because the test statistic -1.2 is in the rejection region of the critical value 1.654. The claim is contained in the alternative hypothesis, so it is supported.
<span>The distri</span>bution coefficient is simply the ratio of the concentrations of the solute in the two solvents.
In this case there is 0.5g of caffeine in 10 ml of chloroform and 0.5 g of caffeine in 100ml of water.
The ratio is 0.5/10 : 0.5/100 = 0.05 : 0.005 = 10 : 1
U = <-6,1>
-4u = -4*<-6,1>
-4u = <-4*(-6),-4*1>
-4u = <24,-4>
----------------------
v = <-5,2>
2v = 2*<-5,2>
2v = <2*(-5),2*2>
2v = <-10,4>
----------------------
-4u + 2v = <24,-4> + <-10,4>
-4u + 2v = <24+(-10),-4+4>
-4u + 2v = <24-10,-4+4>
-4u + 2v = <14,0>
----------------------
Answer: B) <14,0>
Answer:
n = 15
Step-by-step explanation:
For inputs of the value of n, the running time for the algorithm A is 100n^2 and that of B is 2^n.
If A is to run faster than B, 100n^2 must be smaller than 2^n.
Let's check from n = 1 to know the value of n that fits
n = 1
100(1)^2 > 2^1
100 > 2
n = 2
100(2)^2 > 2^2
400 > 4
n = 4
100(4)^2 > 2^4
1600 > 16
n = 8
100(8)^2 > 2^8
6400 > 2^8
n = 16
100(16)^2 < 2^16
25600 < 2^16
This implies that between n = 8 and 16, A starts to run faster than B
n = (8+16)/2 = 12
100(12)^2 > 2^12
14400 > 2^12
n = (12+16)/2 = 14
100(14)^2 > 2^14
19600 > 2^14
n = (14+16)/2
n = 15
100(15)^2 < 2^15
22500 < 2^15
At n= 15, A starts running faster than B
