The distribution of statistics marks of some sandwich students was found to be normal with a mean within a range of 61- 65 and a standard deviation of within a range of 7.8 – 8.6 (1dp). It was found out that there were 348 students who took the course. With an assumed mean and standard deviation, answer the questions that follow: a. If the minimum mark to qualify for an interview was 43. What is the probability that a student selected at random qualified for the interview and what percentage did not qualify? b. If the cut-off mark for selection is 48, how many and what percentage of the students were selected? c. How many students scored a mark greater than 54 but less than 78? d. How many students from the group scored a mark greater than 68 but less than 80?

i really dont know

step-by-step explanation:

ratio is 2 : 9

the answer is three

13.5 : 3

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺