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harina [27]
1 year ago
7

data collected from a coffee shop indicate that the price of a drink forms a consistent pattern that can be graphed as the given

uniform density curve. a graph titled cost of drinks has cost (dollars) on the x-axis. a horizontal line is at y = one-third from 3 to 6. what proportion of drinks cost more than $4.00? one-third one-half two-thirds 1
SAT
1 answer:
ololo11 [35]1 year ago
8 0

The proportion of the drinks having the cost of more than 4 dollars in a coffee shop is two thirds.

<h3>What is a Uniform Density Curve?</h3>

Uniform Density Curve is given as the curve with the rectangle shape defining all the probabilities given for an outcome are identical.

The uniform density curve for the coffee shop have the been plotted as cost of drinks. The x-axis is the representation of the cost, and the plot is formulated one third from 3-6.

From the uniform density curve, the proportion of the drinks that are found to be costing in more than 4 dollars are two third, as one third lies for the cost up to 3 dollars.

Learn more about uniform density curve, here:

brainly.com/question/19254740

#SPJ1

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Which organization would be considered a nonbank institution
inna [77]

The  organization that would be considered a nonbank institution is b) Missouri Public School Retirement System.

<h3>What is nonbank institution?</h3>

A non-banking financial institution  can be regarded as  financial institution that does not have a full banking license .

In this case, The  organization that would be considered a nonbank institution is b) Missouri Public School Retirement System.

CHECK THE COMPLETE QUESTION BELOW:
Which of the following organizations would be considered a nonbank institution?

a) Educational Employees Credit Union

b) Missouri Public School Retirement System

c) Southwest Savings Bank

d) Heartland Bank

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6 0
2 years ago
In the time signature 2/4 the ____ is equal to one beat.
Digiron [165]

Answer: a quater note

Explanation:

just did it mark brainliest plzzz

5 0
2 years ago
Munnabhai knows the four values of agile manifesto by heart.
4vir4ik [10]

Considering the full question, Munnabhai miss to learn <u>Agile Phases</u>.

Given that Munnabhai could not recollect anything about short iterations in Agile Manifesto, it should be noted that iterations is part of <u>Agile Phases</u>.

<h3>What is Agile Phases?</h3>

Agile Phases is the term that is used to describe the stages or phases in which project is executed.

Agile Phases include the following phases:

  • Concept,
  • Inception,
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  • Retirement

Hence in this case, it is concluded that the correct answer is <u>Agile Phases.</u>

The full question is:

Munnabhai knows the four values of Agile Manifesto by heart. However, he was confused when a customer spoke with him, highlighting Agile characteristics of short software development cycles or iterations. He could not recollect anything about short iterations in the Agile Manifesto. What did he miss to learn about?

Learn more about <u>Agile Phases</u> here: brainly.com/question/23661838

5 0
2 years ago
Friction: a 50-kg box is resting on a horizontal floor. A force of 250 n directed at an angle of 30. 0° below the horizontal is
sergejj [24]

The net forces on the box parallel and perpendicular to the surface, respectively, are

∑ F[para] = (250 N) cos(-30.0°) - F[friction] = (50 kg) a

and

∑ F[perp] = F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

where a is the acceleration of the box. (We ultimately don't care what this acceleration is, though.)

To decide whether the friction here is static or kinetic, consider that when the box is at rest, the net force perpendicular to the floor is

∑ F[perp] = F[normal] - F[weight] = 0

so that, while at rest,

F[normal] = (50 kg) g = 490 N

Then with µ[s] = 0.40, the maximum magnitude of static friction would be

F[s. friction] = 0.40 (490 N) = 196 N

so that the box will begin to slide if it's pushed by a force larger than this.

The horizontal component of our pushing force is

(250 N) cos(-30.0°) ≈ 217 N

so the box will move in our case, and we will have kinetic friction with µ[k] = 0.30.

Solve the ∑ F[perp] = 0 equation for F[normal] :

F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

F[normal] - 125 N - 490 N = 0

F[normal] = 615 N

Then the kinetic friction felt by the box has magnitude

F[k. friction] = 0.30 (615 N) = 184.5 N ≈ 185 N

8 0
2 years ago
The sum of two polynomials is modeled below.
marysya [2.9K]

Answer:

hello i can help

Explanation:

its A a square+1

3 0
2 years ago
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