In a fish tank 2/10 of the fish were orange and 5/100 of the fish were stripped.what fraction of the fish were orange or strippe
2 answers:
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Answer:
Hence,we need at least 136 rainfall PH values in the sample i.e
n ≥ 136
Step-by-step explanation:
We are given that:
(σ1)^2 = (σ2)^2 = Population variance = 0.25
So, E < 0.1
Confidence coefficient (c) = 0.9
n = n1 = n2
For confidence level, 1 - α = 0.9,we'll determine Z (α /2) = Z 0.05 by looking up 0.005 using the normal probability table which i have attached.
So, Z (α /2) = 1.645
The margin of error E is given as;
E = Z (α /2)√[(σ1)^2)/n1] + [(σ2)^2)/n2]
= Z (α /2)√({(σ1)^2 + (σ2)^2}/n) < 0.1
Multiply both sides by √n to get;
Z (α /2)√(σ1)^2 + (σ2)^2} < 0.1√n
Divide both sides by 0.1;
{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1 <√n
When we square each side, we get
{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1} ^2 < n
We'll now fill in the known values and solve;
n > ( 1.645 x √{(0.25 + 0.25)/0.1}^2
n > 135.3 or approximately n > 136
Hence,we need at least 136 observations in the sample i.e
n ≥ 136
You should start with an equation.
The 'x' represents the week.
so now, we have to subtract 125 from the total amount which is 245.
now we have
to find 'x', we must divide 15 on each side
we are now left with
that means Sierra will have 8 weeks until she has enough money to buy the bike
The 'x' represents the week.
so now, we have to subtract 125 from the total amount which is 245.
now we have
to find 'x', we must divide 15 on each side
we are now left with
that means Sierra will have 8 weeks until she has enough money to buy the bike