Answer:
The probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5
Step-by-step explanation:
We are given that . At Johnson University, the mean time is 5 hrs with a standard deviation of 1.2 hrs.
Mean = 
Standard deviation = 
We are supposed to find the probability that the average time 100 random students on campus will spend more than 5 hours on the internet i.e. P(X>5)
Z=0
P(X>5)=1-P(X<5)=1-P(Z<0)=1-0.5=0.5
Hence the probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5
.5 is basically 1/2. We know this because .5 written as a fraction is 50/100 which you can simplify to 1/2.
6/10 is greater than 1/2. You can find this out by converting 1/2 to tenths. Multiply the numerator and the denominator by 5 to get 5/10. From there you can easily see that 6/10 > 5/10.
As for the number line simply use 10ths: 1/10, 2/10, 3/10 ... and input 6/10 and 5/10 in the appropriate areas.
If $3 was 25% of the original price then 3*4 = $12
No enough info
you can just say that five brought there own lunch while 19 didnt
Answer:
I don't understand your question but thanks for the points ;) Kinda sucks when someone does it to you, right? Maybe you should stop commenting it on other people's questions so they can get the help they need
