The avage precipitation is 19.32 inches with a standard deviation of 2.44 inches. Find the probability that the precipitation will be greater than 18 inches.

Answers

Answer 1

The probability that the precipitation will be greater than 18 inches is 0.7088.

How to calculate the probability?

From the information, the average precipitation is 19.32 inches with a standard deviation of 2.44 inches.

This will be done using the z score.

Z = (18 - 19.32) / 24

= 0.55

P(X>18) = P(Z> -0.55)

where

P(Z> -0.55) = 1- P(Z<-0.55)

From the normal distribution table

P(Z<-0.55) = 0.2912

Therefore;

P(Z> -0.55)

= 1- 0.2912

= 0.7088

In conclusion, the probability is 0.7088.

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Related Questions

What is the distance δymax−max between the first maxima (on the same side of the central maximum) of the two patterns?

Answers

The distance between consecutive identical points on a wave, such as the space between two succeeding maxima, at a specific point in time is known as the wavelength.

Which laser's first maximum is located more near the center of the beam?The first order maximum generated by laser 1 is therefore closer to the central maximum than the first order maximum generated by laser 2. As can be seen, the distance between two successive interference maxima is constant and regardless of the observed order.The distance between consecutive identical points on a wave, such as the space between two succeeding maxima, at a specific point in time is known as the wavelength.The first order maximum generated by laser 1 is therefore closer to the central maximum than the first order maximum generated by laser 2. As can be seen, the distance between two successive interference maxima is constant and regardless of the observed order.                      

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What was the average weight of a 5’0/4’11 twelve-year-old female in 1977?

please no out out of context or absurd answers! any random answers will be deleted!!

Answers

Answer: The CDC also reports that a 12-year-old girl's weight is usually between 68 and 135 pounds, and the 50th percentile weight for girls is 92 pounds.

Explanation:

Hope its not absurd

The average weight of a 5’0/4’11 twelve-year-old female in 1977 is between 68 and 135 pounds

What is weight?

The force exerted on an object by gravity is known as the weight of the object in science and engineering. The gravitational force acting on the object is referred to as weight in some common textbooks. Some people refer to weight as a scalar quantity that measures the gravitational force's strength.

It gauges how much gravity is pulling on a body. Weight is calculated using the formula w = mg. Since weight is a force, it has the same SI unit as a force, which is the Newton (N). In this case, the weight is about 68 and 135 pounds.

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What do the reactions to sir launcelot's "mad trist" reveal about the narrator's and roderick's different points of view in the passage?

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The narrator overlooks the similarities between Ethelred's acts and those of Lady Madeline, unintentionally aggravating Roderick. This is revealed by the reactions to Sir Launcelot's "Mad Trist," which illustrates the narrator's and Roderick's divergent points of view in the chapter.

The narrator finds the writing of "Mad Trist" uninteresting? But Roderick extrapolates the fictitious action to the storm's happenings, escalating his worries. Mad Trust is a story about intimacy and passion. Roderick added fake action to the story's events in attempt to make it fascinating, despite the narrator's perception that the narrative is dull.The narrator overlooks the similarities between Ethelred's acts and those of Lady Madeline, unintentionally aggravating Roderick. This is revealed by the reactions to Sir Launcelot's "Mad Trist," which illustrates the narrator's and Roderick's divergent points of view in the chapter.The narrator overlooks the similarities between Ethelred's acts and those of Lady Madeline, unintentionally aggravating Roderick.        

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Consider the reaction: mg(oh)₂ (s) ⇌ mg²⁺ (aq) + 2 oh⁻ (aq) at equilibrium, a 1. 0 l reaction vessel contains 5. 3 moles of mg(oh)₂ and concentrations of 0. 0080 m and 0. 010 m of mg²⁺ and oh⁻ respectively. What is kc for this equilibrium? note kc is sometimes called k

Answers

The system's Kc is 8 10-7,The relationship between the equilibrium concentrations of products and reactants.

Finding KC in equilibrium: how do you accomplish it?

Kc Formula: Kc=[C]c[D]d[A]a[B]b] is the formula for calculating Kc. K c = [C] c [D] d [A] a [B] b, where [C] and [D] are the equilibrium molar concentrations of the products and [A] and [B] are the equilibrium molar concentrations of the reactants.

The relationship between the equilibrium concentrations of products and reactants, each raised to the power of their respective stoichiometric coefficients, is known as the equilibrium constant, or Kc.

The reaction's equation is Mg(OH)2 (s) = Mg2+ (aq) + 2 OH (aq)

The question instructs us to look for the Kc of the reaction.

Kc = [Mg2+] [OH]

Remember that since Mg(OH)2 is a pure solid, it is not included in the equilibrium expression.

substituting values;

[Mg²⁺] = 0.0080 M

[ OH⁻ ] = 0.010 M

Kc = [0.0080 M] [0.010 M ]^2

Kc = 8 × 10^-7.

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A monkey has filled in a 3 × 3 grid with the numbers 1, 2,. . . , 9. A cat writes down the three numbers obtained by multiplying the numbers in each horizontal row. A dog writes down the three numbers obtained by multiplying the numbers in each vertical column. Can the monkey fill in the grid in such a way that the cat and dog obtain the same lists of three numbers? what if the monkey writes the numbers 1, 2,. . . , 25 in a 5 × 5 grid? or 1, 2,. . . , 121 in a 11 × 11 grid? can you find any conditions on n that guarantee that it is possible or any conditions that guarantee that it is impossible for the monkey to write the numbers 1, 2,. . . , n2 in an n × n grid so that the cat and the dog obtain the same lists of numbers?

Answers

Yes, it is possible for the monkey to fill in the 3 x 3 grid in such a way that the cat and dog obtain the same lists of three numbers.

What if the monkey writes the numbers 1, 2,. . . , 25 in a 5 × 5 grid? or 1, 2,. . . , 121 in a 11 × 11 grid?For example, the monkey could fill the 3x3 grid with the numbers 1, 4, 7; 2, 5, 8; and 3, 6, 9.Yes, it is possible for the monkey to fill in the 5 x 5 grid with the numbers 1, 2, ... , 25 in such a way that the cat and dog obtain the same lists of three numbers. For example, the monkey could fill the 5x5 grid with the numbers 1, 8, 15, 22, 25; 2, 7, 16, 21, 24; 3, 6, 17, 20, 23; 4, 5, 18, 19, 12; and 9, 10, 11, 14, 13.Yes, it is possible for the monkey to fill in the 11 x 11 grid with the numbers 1, 2, ... , 121 in such a way that the cat and dog obtain the same lists of three numbers. For example, the monkey could fill the 11x11 grid with the numbers 1, 10, 19, 28, 37, 46, 55, 64, 73, 82, 121; 2, 9, 20, 27, 38, 45, 56, 63, 74, 81, 120; 3, 8, 21, 26, 39, 44, 57, 62, 75, 80, 119; 4, 7, 22, 25, 40, 43, 58, 61, 76, 79, 118; 5, 6, 23, 24, 41, 42, 59, 60, 77, 78, 117; 11, 12, 13, 14, 15, 16, 17, 18, 33, 34, 115; 10, 13, 22, 25, 34, 37, 46, 49, 68, 71, 114; 9, 14, 21, 26, 33, 38, 45, 50, 67, 72, 113; 8, 15, 20, 27, 32, 39, 44, 51, 66, 73, 112; and so on.In general, it is possible for the monkey to fill in an n x n grid with the numbers 1, 2, ... , n2 in such a way that the cat and dog obtain the same lists of numbers, provided that n is an even number. For odd numbers (n = 3, 5, 7, etc.), it is impossible for the monkey to fill in the grid in such a way that the cat and dog obtain the same lists of numbers.

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Mariela is standing in a building and looking out of a window at a tree. The tree is 2020 feet away from mariela. Mariela's line of sight to the top of the tree creates a 42^{\circ}42 ∘ angle of elevation, and her line of sight to the base of the tree creates a 31^{\circ}31 ∘ angle of depression

Answers

30.01 feet is the tree's height, the distance between the tree's base and its highest point on the ground is known as the total tree height.

How would one characterize the size of a tree?A DBH of up to 14 inches is considered little, 15–19 inches is considered medium, and 20 inches or more is considered large. Caliper, a measurement of a tree's diameter used in the nursery industry that takes into account the size of the plant at a point 6 to 12 inches above the ground, is how we characterize trees.The distance between the tree's base and its highest point on the ground is known as the total tree height. The use of a yardstick is a quick and fairly accurate way to gauge the height of a tree.30.01 feet is the tree's height, the distance between the tree's base and its highest point on the ground is known as the total tree height.

The complete question is,

A tree may be seen through a window where Maria is standing within a structure. The distance between Mariela and the tree is 20 feet; the angle of elevation and depression created by Mariela's line of sight are 42 degrees and 31 degrees, respectively. Describe the tree's height in feet.

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A triangle has sides with lengths of 7 meters, 14 meters, and 17 meters. Is it a right triangle?

Answers

No, it is not a right angled  triangle.

For a right angled triangle, Acc. to Pythagoras theorem:-

Hypotenuse²= Base² + Perpendicular²

Hypotenuse is the longest side of the triangle.

For the given triangle, Hypotenuse= 17

17² ≠ 14² + 7²

289 ≠245.

Hence, it is not a right angled triangle.

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What is the radial electric field at the point r = 0. 5(r1+r2)? give your answer in units of kn/c

Answers

The radial electric field is -13.8 kN/C at the point where r is 0.5(R1+R2) and is +8.51 kN/C at the point where r is 2R2.

Gaussian law?

The ratio of the total charge contained within this surface to the electric permittivity determines the electric field through the hypothetical surface. a made-up surface encircling a charge distribution. The ratio of the total charge contained within this surface to the electric permittivity determines the electric field through the hypothetical surface. a made-up surface encircling a charge distribution.

Between the sphere's inner and outer shells, a fictitious surface has been drawn in (A).

Charge = Q.

r= 0.5(R1+R2) = 12 cm.

(B) If an imagined surface is drawn outside of the shell,

As a result, the radial electric field at point r=0.5(R1+R2) is -13.8 kN/C, whereas the radial electric field at point r=2R2 is +8.51 kN/C.

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