Answer: 0.0107
Step-by-step explanation:
Given : The time for the event for boys in secondary school is known to possess a normal distribution with
Let x be the random variable that represents the time for mile run in its secondary-school fitness test.
z-score :
Now, the probability that a randomly selected boy in secondary school can run the mile in less than 322 seconds is given by :-
Hence, the probability that a randomly selected boy in secondary school can run the mile in less than 322 seconds = 0.0107
Answer:
Divergent
Step-by-step explanation:
an = 3 sin(2/n)
Choose bn = 3 (2/n).
Using Limit Comparison Test:
lim(n→∞) an / bn
= lim(n→∞) [3 sin(2/n)] / [3 (2/n)]
= lim(n→∞) [sin(2/n)] / (2/n)
= 1
The limit is greater than 0, and bn diverges, so an also diverges.
Yes at 0.05 significance level we can conclude that mean income of those selecting plan B is bigger.
Given mean yearly income for sample of 40 subscribers is $57000 with standard deviation of $9200, mean yearly income for sample of 30 subscribers with standard deviation of $7100.
First we have been given significance level (α) of 0.05. So (1-α)=1-0.05=0.95 and we have to find Z score for this significance level. (1+0.95)/2=0.425
Z=1.44
Now we have to calculate margin of error which will be calculated as under:
=
=1.44*9200/6.32
=1869.10
=
=1.44*7100/3.47=2096.2
We know that margin of error shows the difference between real and calculated values. So if we increase mean of both plans by margin of error of respective plans then mean of plan A becomes=57000+1869.10=58869.10 and mean of plan B becomes =61000+2096.2=630622.2. We can clearly notice that mean of plan B is greater.
Hence it is reasonable to say that the mean of plan B is greater at the significance level of 0.05.
Learn more about margin of error at
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The second one is the correct answer. Plug in -3 into the equation.
<u>x-3</u>
x+7
its suppose to be at fraction