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alina1380 [7]
3 years ago
7

Suppose the dealer incentive per vehicle for honda's acura brand in 2012 is thought to be bell-shaped and symmetrical with a mea

n of $2500 anda standard deviation of $300. Based in this information,
what interval of dealer incentives would we expect approximately 99.7% of vehicles to fall within?
Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

From $1600 to $3400.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 2500

Standard deviation = 300

What interval of dealer incentives would we expect approximately 99.7% of vehicles to fall within?

By the Empirical Rule, 99.7% fall within 3 standard deviations frow the mean. So

From 2500 - 3*300 = 1600 to 2500 + 3*300 = 3400.

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The mean of the temperatures in the chart is 24° with standard deviation of 4°. How many years had temperatures within one stand
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Answer:

Step-by-step explanation:

Given: The mean of the temperatures in the chart is 24° and standard deviation is 4°.

To find: How many years had temperatures within one standard deviation of the mean

Solution: Mean temperature μ = 24 , standard deviation σ =4

The lower and upper bound for temperature within one standard deviation of the mean is given as:

Lower bound = μ - σ = 24 - 4=20

Thus, the Lower bound is = 20

Upper bound = μ + σ = 24+ 4=28

Thus, the Upper bound is = 28

Now, The temperature value between (Lower bound, Upper bound) that is (20, 28) is said to be within one standard deviation of the mean.

Hence 27° is the temperature value that is within one standard deviation of mean.

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Help meez 40 pts use surface area formula of cylinder that is for Lateral surface area and for total surface area
jeka94

Answer:So the radius of the cylinder is 2.65 cm.

A cylinder can be defined as a solid figure that is bound by a curved surface and two flat surfaces. The surface area of a cylinder can be found by breaking it down into 2 parts:

1.  The two circles that make up the caps of the cylinder.

2.  The side of the cylinder, which when "unrolled" is a rectangle.

The area of each end cap can be found from the radius r of the circle, which is given by:

A = πr2

Thus the total area of the caps is 2πr2.

The area of a rectangle is given by:

A = height × width

The width is the height h of the cylinder, and the length is the distance around the end circles, or in other words the perimeter/circumference of the base/top circle and is given by:

P = 2πr

Thus the rectangle's area is rewritten as:

A = 2πr × h

Combining these parts together we will have the total surface area of a cylinder, and the final formula is given by:

A = 2πr2 + 2πrh

where:

π  is Pi, approximately 3.142

r  is the radius of the cylinder

h  height of the cylinder

By factoring 2πr from each term we can simplify the formula to:

A = 2πr(r + h)

The lateral surface area of a cylinder is simply given by: LSA = 2πr × h.

Example 1: Find the surface area of a cylinder with a radius of 4 cm, and a height of 3 cm.

Solution:

SA = 2 × π × r2 + 2 × π × r × h

SA = 2 × 3.14 × 42 +  2 × 3.14 × 4 × 3

SA = 6.28 × 16 + 6.28 × 12

SA = 100.48 + 75.36

SA = 175.84

Surface area = 175.84 cm2

Example 2: Find the surface area of the cylinder with a radius of 5.5cm and height of 10cm.

Solution:

The radius of cylinder = 5.5 cm.

The height of cylinder = 10 cm.

The total surface area of the cylinder is therefore:

TSA = 2πr(r+h)

TSA = 11π (5.5+10)

TSA = 170.5 π

TSA = 535.6 cm2

Example 3: Find the total surface area of a cylindrical tin of radius 17 cm and height 3 cm.

Solution:

Again as in the previous example:

TSA = 2πr(r+h)

TSA = 2π× 17(17+3)

TSA = 2π×17×20

TSA = 2136.56 cm2

Example 4: Find the surface area of the cylinder with radius of 6 cm and height of 9 cm.

Solution:

The radius of cylinder: r = 6 cm

The height of cylinder: h = 9 cm

Total surface area of cylinder is therefore:

TSA = 2πr(r + h)

TSA = 12π (6+9)

TSA = 180 π

TSA = 565.56 cm2

Example 5: Find the radius of cylinder whose lateral surface area is 150 cm2 and its height is 9 cm.

Solution:

Lateral surface area of cylinder is given by:

LSA = 2πrh

Given that:

LSA = 150cm2

h = 9cm

π is the constant and its value = 3.14

Substitute the values in the formula and find the value of r by isolating it from the equation:

LSA = 2πrh

150 = 2× π × r × 9

r = 150 / (2×9× π)

r = 2.65cm

So the radius of the cylinder is 2.65 cm.

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2 years ago
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