Assuming that a soap bubble is a perfect sphere, what is the diameter of a bubble containing 1,000 cm3 of air? Round to the near
1 answer:
Answer:
17.85 cm
Step-by-step explanation:
1) Area of a sphere formula:
2) Plug in A: 1000 =
3) Simplify: 1000 = 12.56
4) Divide Each Side By 12.56: 79.61783439 =
5) Take the Square Root of 79.61783439 to Solve for : 8.922882628 =
6) Double to get the diameter: 8.922882628*2= 17.84576486
7) Round to the Nearest Tenth: 17.85 cm
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Answer:
n= 60
Step-by-step explanation:
Hello!
You have Y₁, Y₂, ..., Yₙ random sample with a normal distribution: Y~N(μ;σ²)
μ= 2
σ²= 4
You need to calculate a sample size n so that (1.9 ≤ Y ≤2.1)= 0.99
To reach the sample size you need to work with the distribution of the sample mean (Y[bar]) because it is this distribution that is directly affected by the sample size.
Y[bar]~N(μ;σ²/n)
Under the sample mean distribution you have to use the standard normal:
Z= Y[bar] - μ ~N(0;1)
σ/√n
Now the asked interval is:
P(1.9 ≤ Y[bar] ≤2.1)= 0.99
The upper bond is 2.1
The lower bond is 1.9
The difference between the two bonds is the amplitude of the interval a=2.1-1.9= 0.2
And the probability included between these two bonds is 0.99
With this in mind you can rewite it as an interval for the sample mean:
Y[bar] + *(σ/√n) - (Y[bar] +
*(σ/√n))= 0.2
Using the semiamplitude (d) of the interval you can easly calculate the required sample:
d= a/2= 0.2/2= 0.1
d= *(σ/√n)
d* = σ/√n
√n*(d* [tex]Z_{1-\alpha /2}[/tex)= σ
√n= σ/(d* [tex]Z_{1-\alpha /2}[/tex)
n= (σ/(d* [tex]Z_{1-\alpha /2}[/tex))²
n= (2/(0.1* 2.586))²
n= 59,81 ≅ 60
I hope it helps!