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.
9 - substitute y for 6, then subtract that from 36, which is 18. 18 divided by 2 is 9
That would be a scalene triangle, because all three sides are different measurements. Equilateral means all 3 sides are equal and isocelese means that 2 sides are equal. Therefore it would have to be scalene
Y=3x2 3x(0.5)=1.5+2=3.5 Y=3.5
3(0.5)X2 1.5X2=3 Y=3
1.5/2=0.75 Y=0.75
1.5-2=-0.5 Y= -0.5
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