1/8 inch thick
inner diameter of 8.5

volume shell=volume total-volume shell

1/8=0.125

diameter/2=radius
radius inner=8.5/2=4.25
radiustotal=inner+outer=4.25+0.125=4.375

so

Vouter-Vinner=(4/3)pi4.375³-(4/3)pi4.25³
Vshell=(4/3)(pi)(4.375³-4.25)
Vshell=(4/3)(pi)(83.740234375-76.765625)
Vshell=(4/3)(pi)(6.974609375)
Vshell=9.299479166666666666666pi cubic inches
Vshell≈16.433536180619 cubic inches

6 0

2 years ago

Read 2 more answers

Suppose a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg,

lianna [129]

Answer:

(a) Probability that Y falls into the dangerous region is 0.0013.

(b) Probability that the mean Y-bar falls into the dangerous region is 0.00001.

Step-by-step explanation:

We are given that a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg, σ = 1 mg).

A dosage of 13 mg is considered dangerous.

Let Y = dosage of the active ingredient

The z-score probability distribution for normal distribution is given by;

Z = $\frac{ Y-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = population mean = 10 mg

$\sigma$ = standard deviation = 1 mg

(a) Probability that Y falls into the dangerous region is given by = P(Y $\geq$ 13 mg)

P(Y $\geq$ 13 mg) = P( $\frac{ Y-\mu}{\sigma} } }$ $\geq$ $\frac{ 13-10}{1} } }$ ) = P(Z $\geq$ 3) = 1 - P(Z < 3)

= 1 - 0.9987 = 0.0013

The above probability is calculated by looking at the value of x = 3 in the z table which has an area of 0.9987.

(b) We are given that a dosage of 13 mg is considered dangerous. And we sample 49 capsules at random.

Let $\bar Y$ = sample mean dosage

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }$ ~ N(0,1)

where, $\mu$ = population mean = 10 mg

$\sigma$ = standard deviation = 1 mg

n = sample of capsules = 49

So, Probability that the mean Y-bar falls into the dangerous region is given by = P( $\bar Y$ $\geq$ 13 mg)

P(Y $\geq$ 13 mg) = P( $\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }$ $\geq$ $\frac{13-10}{\frac{1}{\sqrt{49} } } } }$ ) = P(Z $\geq$ 21) = 1 - P(Z < 21)

= 0.00001

6 0

2 years ago