Answer:
EG/LN similar FG/MN
Step-by-step explanation:
Answer:
Step-by-step explanation:
The mean SAT score is
, we are going to call it \mu since it's the "true" mean
The standard deviation (we are going to call it
) is

Next they draw a random sample of n=70 students, and they got a mean score (denoted by
) of 
The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.
- So the Null Hypothesis 
- The alternative would be then the opposite 
The test statistic for this type of test takes the form

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.
With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>
Answer:
3x^3-2x^2-6x+4
Step-by-step explanation:
Your welcome!!!!!!!!!!!!
Ax+b=-6x + 36
a is -6 and b is 46
Let us calculate the speed of each conveyor belt. The bigger one does 300 pounds in 10 minutes, hence its speed is 300 pounds/ 10 minutes=30 pounds/min. The smaller one needs 30 minutes, hence its speed is 300 pounds/ 30 minutes=10 pounds/min. We see that the total speed of both machines if they work together is 30+10=40 pounds/min. We have then that the two machines together can transfer 300 pounds in 300/40=7.5 minutes.