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2 years ago

Consider a single spin of the spinner.

Mathematics

2 answers:

alex41 [277]2 years ago

6 0

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed. These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

Cloud [144]2 years ago

5 0

Answer:

2nd and 5th Option

2nd Option: Landing on a shaded portion and landing on a number greater than 3

5th Option: Landing on an unshaded portion and landing on a number less than 2

Step-by-step explanation:

edge 2020

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A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0

2 years ago

The area of a sign in the shape of a square is 219 square inches. Which measurement is the closest to the side length of this si

Aloiza [94]

Answer:

Step-by-step explanation:

7 0

2 years ago

Complete the sentence below. If (3x2 + 22x + 7) ÷(x + 7) = 3x + 1, then (x + 7)(???????) =???? .

g100num [7]

Answer: The answer is $(x+7)(3x+1)=33x^2+22x+7.$

Step-by-step explanation: Given that

$(3x2 + 22x + 7) \div(x + 7) = 3x + 1,$

and we are to complete the following sentence:

$(x+7)(???)=???$

We have the following division algorithm for polynomials

$\textup{If }a(x)\times b(x)=c(x),\\\textup{then, we have }\\\\\dfrac{c(x)}{b(x)}=a(x)~~~~~\textup{or}~~~~~~c(x)\div b(x)=a(x).$

Here, a(x) = quotient, b(x) = divisor and c(x) = dividend.

Applying this rule in the given problem, we have

$\textup{since }(3x2 + 22x + 7) \div(x + 7) = 3x + 1,\\\\\textup{so, }\\\\(x+7)(3x+1)=3x^2+22x+7=0.$

Thus, the complete sentence is

$(x+7)(3x+1)=3x^2+22x+7=0.$

7 0

2 years ago

Read 2 more answers

The given equation has been solved in the table. In which step was the subtraction property of equality applied?

atroni [7]

Answer:

Option (D)

Step-by-step explanation:

Subtraction property of equality tells that whatever subtracted from one side of the equation must be subtracted from the other side.

If x + 2 = 2,

By the property of subtraction of equality,

x + 2 - 2 = 2 - 2

x = 0

But in the given question,

$\frac{x}{2}-7=-7$

$\frac{x}{2}-7+7=-7+7$

shows the addition property of equality in step (2)

Therefore, subtraction property of equality was not applied.

Option (D) will be the answer.

5 0

1 year ago

Graph a line with a slope of -2/5 that contains the point (-3,5)

kotykmax [81]

$y=-\frac{2}{5}x+\frac{19}{5}$

Further explanation:

We have to find the equation of the line first to graph the line.

The general form of slope-intercept form of equation of line is:

$y=mx+b$

Given

$m=-\frac{2}{5}$

Putting the value of slope in the equation

$y=-\frac{2}{5}x+b$

To find the value of b, putting the point (-3,5) in equation

$5=-\frac{2}{5}(-3)+b\\5=\frac{6}{5}+b\\5-\frac{6}{5}+b\\b=\frac{25-6}{5}\\b=\frac{19}{5}$

Putting the values of b and m

$y=-\frac{2}{5}x+\frac{19}{5}$

The graph is made by using online graphing tool Desmos.

Keywords: Equation of line, graph of line

Learn more about graphs at:

#LearnwithBrainly

4 0

2 years ago