Answer:
H0 rejected as μ>4
and a fish is unsafe to eat from the lake.
Step-by-step explanation:
1)Let the null hypothesis be H0 : mu ≤4 against the alternate Ha: mu > 4 ppb
H0: μ ≤4 against the claim Ha: μ>4
2) X`= ∑x/ n= 2.9 +7.6+ 4.8+ 5.2+ 5.1+ 4.7+ 6.9+ 4.9+ 3.7+ 3.8/10
= 49.6/10= 4.96
The standard deviation can be calculated sigma= 1.344767 ( using statistic calculator)
3) The significance level is taken to be ∝=0.05
The value of z at 0.05 for 1 sided test is z >± 1.645
i.e the critical region is less than - 1.645 and greater than +1.645
4) Taking the distribution to be approximately normal
Z= x`- u / s/ √n
Z= 4.96-4/ 1.345/10
Z= 4.96-4/ 1.345/3.1622
Z= 0.96/0.4253
Z= 2.257
5) Since the calculated value of z = 2.257 falls in the critical region we reject our null hypothesis and conclude that true mean value of the PCB concentration is greater than 4 (ppb) and that a fish is unsafe to eat from the lake.
Answer:
20 sides.
Step-by-step explanation:
The exterior angle = 180 - 162 = 18 degrees.
Since the sum of the exterior angles = 360 degrees the number of sides
= 360 / 18
= 20 sides.
Answer:1/6
Step-by-step explanation:
If you graph it just count how many your line pushes over than down from whole numbers, and that should be your slope (sorry if it’s wrong haven’t don slope in a million years)
1) He worked 4 hours leaving 6 more hours for him.
rate for A alone 6 baker
rate for A & B together 4
1/a+1/b=1/x
1/6+1/b=1/4
1/b=1/4-1/6
1/b=6/24-4/24
1/b=2/24
1/b=1/12
b=12
rate for wife alone 12 hours for the remaining 6 hours
check
1/6+1/12=1/4
12/72+6/72=1/4
18/72=1/4
ok
codewa
But we want to know how long it would have taken her for the whole job not just part of it
he works twice as fast as she does so the job that takes him 10 hours alone would take her 20 hours alone.
2) jessica and cynthia work in a pet shop.
it takes jessica 6 hours to groom all the pets but cynthia needs 8 hours to groom them.
if jessica starts to groom 1 hour before cynthia joins, how long will it take them to finish?
:
let t = time it takes them to finish (C's working time)
then
t+1 = J's working time
let 1 = completed job (all pets groomed)
:
A shared work equation, each does a fraction of the job, the two fractions add up to 1
%28%28t%2B1%29%29%2F6 + t%2F8 = 1
Multiply by the least common multiple of 6 and 8, 24. cancel the denominators
4(t+1) + 3t = 24
4t + 4 + 3t = 24
4t + 3t = 24 - 4
7t = 20
t = 20/7
t = 26%2F7 hrs, which is: 2 + 6%2F7*60 = 2 hrs 51.43 min to finish the job
3) math club takes 40 minutes
science takes 50 minutes.
rate * time = quantity
quantity = 1 set up of the tables.
rate for math club is 1/40 of all the tables in 1 minute.
rate for science club is 1/50 of all the tables in 1 minute.
math club works for 10 minutes.
rate * time = quantity
1/40 * 10 = 10/40 = 1/4 of the tables are already set up by the math club.
there are 3/4 of the tables that still need to be set up.
rate * time = quantity
the rates of the math club and the science club are additive.
(1/40 + 1/50) * time = 3/4 of the tables that still need to be set up.
common denominator is 200.
(5/200 + 4/200) * time = 3/4.
9/200 * time = 3/4
time = 3/4 * 200/9 = 600 / 36 = 16 and 2/3 minutes.
total time required is 10 + 16 + 2/3 minutes = 26 + 2/3 minutes.
how to check.
10 minutes is used to do 1/4 of the tables.
16 + 2/3 minutes is used to do the remaining 3/4 of the tables.
combined rate of math and science club is 9/200 of the tables in 1 minute.
math club rate of 1/40 * 10 = 10/40 = 1/4 of the tables.
combine rate of 9/200 * (16 + 2/3) = 9/200 * 50/3 = 450/600 = 9/12 = 3/4 of the tables.
1/4 + 3/4 = 1 = all of the tables.
Which one is it the top one or the bottom one?
