Do you know. How to solve these?
2 answers:
B, e, and c
this is how you can tell
b: 7/2, both the top and bottom × 3 will get 21/6, 7×3=21, 2×3=6 so they are proportional
e: both 5 and 2 multiply 5: 5×5=25, 2×5=10, so 25/10 is proportional to 5/2
c 4×3=12, 5×3=15
the others are not.
another way to do it is what I call 'diagonal times'. for example;
, you multiply diagonally, 5×11=55, 6×9=54, 55 and 54 are not equal, so they two fractions are not proportional.
try this "diagonal times" on b, c, e, you'll see you get two products that are equal.
this is how you can tell
b: 7/2, both the top and bottom × 3 will get 21/6, 7×3=21, 2×3=6 so they are proportional
e: both 5 and 2 multiply 5: 5×5=25, 2×5=10, so 25/10 is proportional to 5/2
c 4×3=12, 5×3=15
the others are not.
another way to do it is what I call 'diagonal times'. for example;
try this "diagonal times" on b, c, e, you'll see you get two products that are equal.
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Answer:
a) 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!
b) 1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!
c) ∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³
Step-by-step explanation:
Given that;
if n ⇒ ∞
p ⇒ 0
⇒ np = Constant = λ, we can apply poisson approximation
⇒ Here 'p' is small ( p=0.01)
⇒ if (n=large) we can approximate it as prior distribution
⇒ let the number of defective items be d
so p(d) = ((e^-λ) × λ) / d!
NOW
a)
Let there be x number of repairs, So they will repair 20x machines on time. So if the number of defective machine is greater than 20x they can not repair it on time.
λ[n0.01]
p[ d > 20x ] = 1 - [ d ≤ 20x ]
= 1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k!
b)
Similarly in this case if number of machines d > 80x/3;
Then it can not be repaired in time
p[ d > 80x/3 ]
1 - ∑^(80x/d)_k=0 ((e^-λ) × λ^k) / k!
c)
n = 300, lets do it for first case i.e;
p [ d > 20x } ≤ 0.01
1 - ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.01
⇒ ∑²⁰ˣ_k=0 ((e^-λ) × λ^k) / k! = 0.99
⇒ ∑²⁰ˣ_k=0 (λ^k)/k! = 0.99e^λ
∑²⁰ˣ_k=0 (3^k)/k! = 0.99e³
Answer:
The 95% confidence interval obtained with a sample size of 64 will give greater precision.
Step-by-step explanation:
We are given the following in the question:
A 95% confidence interval is calculated with the following sample sizes
The population mean and standard deviation are unknown.
Effect of sample size on confidence interval:
- As the sample size increases the margin of error decreases.
- As the margin of error decreases the width of the confidence level decreases.
- Thus, with increased sample size the width of confidence level decreases.
If we want a confidence interval with greater precision that is smaller width, we have to choose the higher sample size.
Thus, the 95% confidence interval obtained with a sample size of 64 will give greater precision.