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rosijanka [135]
3 years ago
14

The machine described has standard deviation σ that can be fixed at certain levels by carefully adjusting the machine. What is t

he largest value of σ that will allow the actual amount dispensed to fall within 1 ounce of the mean with probability at least .95?
Mathematics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

σ = 0.5102

Step-by-step explanation:

If

Y = volume filled,  

µ = mean

It is a normal distribution

It follows that

0.95 = P (|Y − µ| < 1) = P (|Z| <  1 / σ)

so that

1 / σ = 1.96 ⇒  σ = (1 / 1.96)

⇒  σ = 0.5102.

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The image of the triangle having these measurements is attached below:

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Using Pythagorean theorem, we have,

tan \ 45^{\circ}= \frac{AC}{3}

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Thus, the length of AC is $A C=3 \ in

Therefore, Option A is true about the triangle.

Hence, Option A is the correct answer.

Option B: $B C=3\ in

Using Pythagorean theorem, we have,

cos \ 45^{\circ}=\frac{3}{BC}

     BC=\frac{3}{cos \ 45^{\circ}}

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Thus, the length of BC is BC=4.24 \ in

Therefore, Option B is not true about the triangle.

Hence, Option B is not the correct answer.

Option C: $B C=6\ in

Using Pythagorean theorem, we have,

cos \ 45^{\circ}=\frac{3}{BC}

     BC=\frac{3}{cos \ 45^{\circ}}

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     BC=4.24

Thus, the length of BC is BC=4.24 \ in

Therefore, Option C is not true about the triangle.

Hence, Option C is not the correct answer.

Option D: $A C=6\ in

Using Pythagorean theorem, we have,

tan \ 45^{\circ}= \frac{AC}{3}

   1\times 3= AC

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Thus, the length of AC is $A C=3 \ in

Therefore, Option D is not true about the triangle.

Hence, Option D is not the correct answer.

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