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: 600 people
Step-by-step explanation:
That is because if 6,000 people voted, and 10% were those ages, you would have to find what 10% of 6,000 was. Well, 10 has 1 zero and 6,000 has 3, so take off one zero from both number and you get 600 and 1. 600x1= 600.
Answer:
90
Step-by-step explanation:
each dot = 45 bc 540 / how many dots there are which is 12 = 45 then there are 2 dots worth of 45 making it 90
math
540/12=45
45 x 2 = 90
<h3>The answer to your question is k (-3) = 21!</h3>
Here's how I got this answer:
<em><u>K(a) = 2a^2 - a </u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)</u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3</u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3K (-3) = 18 + 3</u></em>
<em><u>K(a) = 2a^2 - a K(-3) = 2(-3)^2 - (-3)K(-3) = 2(9) + 3K (-3) = 18 + 3K (-3) = 21</u></em>
I hope this helps!
Also sorry for the late answer, I just got the notification that you replied to my comment, I hope I came in time!
