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Natalija [7]
3 years ago
13

A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies

t 5% of fruits weigh more than how many grams?
Mathematics
1 answer:
RoseWind [281]3 years ago
8 0

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

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

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

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

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JulijaS [17]

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2 years ago
The cereal box shown below is a rectangular prism. Find the surface area of the cereal box.
aksik [14]

Answer:

Surface area is found:

Surface Area = 1700 cm²

Step-by-step explanation:

(The cereal box is shown in the ATTACHMENT)

The surface area of a rectangular prism can be found by added the areas of all 6 sides of the rectangular prism.

L = length = 20 cm

H = height = 30 cm

W = Width = 5 cm

<h3 /><h3>Side 1:</h3>

A(1) = L×H

A(1) = 20×30

A(1) = 600 cm²

<h3>Side 2:</h3>

As the measurements of the side at the back of side 1 has the same measurement of side 1. then:

A(2) = 600 cm²

<h3>Side 3:</h3>

A(3) = L×W

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<h3>Side 4:</h3>

As the measurements of the side at the back of side 4 has the same measurement of side 4. then:

A(4) = 100 cm²

<h3>Side 5:</h3>

A(5) = H×W

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A(5) = 150 cm²

<h3>Side 6:</h3>

As the measurements of the side at the back of side 5 has the same measurement of side 5. then:

A(6) = 150 cm²

<h3>Surface Area:</h3>

Adding areas of all the sides

A(1) + A(2) + A(3) +A(4) + A(5) + A(6) = 600 + 600 + 100 +100 + 150 +150

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4 years ago
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3/4 would be a fraction, equivalent to 6/8.
4 0
3 years ago
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F(x)=1 |12.0-x<br> F(x)=0.2
Snezhnost [94]

Answer:

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Step-by-step explanation:

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3 years ago
Mr. Brady is using a coordinate plane to design a treasure hunt for his students. The hunt begins at the flagpole. The first clu
dmitriy555 [2]

Answer:

The figure is attached and the total distance is 1031 feet.

Step-by-step explanation:

The graph is indicated in the attached figure.

For calculation of distance consider following

Point Flagpole is (0,0)

Point Clue1 is (0,5)

Point Clue2 is (6,0)

Point Clue3 is (0,-5)

So the distance is calculated as follows

d_{T}=[d_{FP\ to\ Clue1}+d_{Clue1\ to\ Clue2}+d_{Clue2\ to\ Clue3}]*distance\ per\ unit\\d_{T}=[\sqrt{(Clue1_x-FP_x)^2+(Clue1_y-FP_y)^2}+\sqrt{(Clue2_x-Clue1_x)^2+(Clue2_y-Clue1_y)^2}+\sqrt{(Clue3_x-Clue2_x)^2+(Clue3_y-Clue2_y)^2}]**distance\ per\ unit\\Substituting the values

d_{T}=[\sqrt{(0-0)^2+(5-0)^2}+\sqrt{(6-0)^2+(0-5)^2}+\sqrt{(0-6)^2+(-5-0)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{(0)^2+(5)^2}+\sqrt{(6)^2+(-5)^2}+\sqrt{(-6)^2+(-5)^2}]*50 \text{ feet}\\d_{T}=[\sqrt{0+25}+\sqrt{36+25}+\sqrt{36+25}]*50 \text{ feet}\\d_{T}=[\sqrt{25}+\sqrt{61}+\sqrt{61}]*50 \text{ feet}\\d_{T}=[5+7.81+7.81]*50 \text{ feet}\\d_{T}=[20.62]*50 \text{ feet}\\d_{T}=1031 \text{ feet}\\

So the total distance travelled is 1031 feet.

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