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elena-14-01-66 [18.8K]
3 years ago
9

"The weight of a product is normally distributed with a mean of four ounces and a variance of .25 squared ounces. What is the pr

obability that a randomly selected unit from a recently manufactured batch weighs no more than 3.5 ounces?
Mathematics
1 answer:
insens350 [35]3 years ago
5 0

Answer:

Step-by-step explanation:

Since the weight of the products are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = weight of the product.

µ = mean weight

σ = standard deviation

From the information given,

µ = 4 ounces

Variance = 0.25²

σ = √variance = √0.25² = 0.25

We want to find the probability that a randomly selected unit from a recently manufactured batch weighs no more than 3.5 ounces. It is expressed as

P(x ≤ 3.5)

For x = 3.5,

z = (3.5 - 4)/0.25 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.0228

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Marielle’s painting has the dimensions shown (Painting Width = 18 inches, Painting Length = 24 inches). The school asks her to p
monitta

Answer

Find out the  what is the constant proportionality .

To prove

Formula

Area of rectangle = Length × Breadth

As given

Marielle’s painting has the dimensions shown (Painting Width = 18 inches, Painting Length = 24 inches).

Area of the original painting = 18 × 24

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As given

The school asks her to paint a larger version that will hang in the cafeteria.

The larger version will be twice the width and twice the height.

Length of the  larger version painting = 2 × 18

                                                              = 36 inches

Breadth of the  larger version painting = 2 × 24

                                                                = 48 inches

Now area of the larger version painting = 36 × 48

                                                                  = 1728 inches²

As given

The  area of the original painting proportional to the area of the larger

painting .

Thus

Area\ of\ original\ painting\ \propto\ Area\ of\ larger\ painting

Area\ of\ original\ painting\ = k\times \ Area\ of\ larger\ painting

Where k is the constant  of proportionality

Putting the value

432 = k × 1728

k = \frac{1728}{432}

k = 4

The constant proportionality is 4 .

Hence proved  







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