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.
15/16 = 0.9375
90% = 0.9
0.925 = 0.925
7/8 = 0.875
.92 = .92
0.875, 0.9, 0.92, 0.925, 0.9375
least to greatest = 7/8, 90%, .92, 0.925, 15/16
3 2/3 = 3.666666666666667
362% = 36.2
3.66 = 3.66
3 3/5 = 3.6
36
3.6, 3.66, 3.666666666666667, 36, 36.2
3 3/5, 3.66, 3 2/3, 36, 36.2
Answer:d
Step-by-step explanation:
12,000 x 6.5% = 780 divided by 2 =390
6/5=6+0.5=6+½+=6½
6½ is the answer
