52 POINTS, WILL GIVE BRAINLIEST!


Use the Polygon tool to draw a rectangle with a length of 6 units and a height of 4 units. One of the sides of the rectangle falls on line CD , and the rectangle has a vertex of C.


Each segment on the grid represents 1 unit.

Answer: $ 22,466.45

Step-by-step explanation:

Given : Future value : [tex]FV= \$50,000[/tex]

The number of time period : [tex]t=10\text{ years}[/tex]

The rate of interest : [tex]r=8\ %=0.08[/tex]

Let P be the present value.

The formula to calculate the future value is given by :-

[tex]FV=Pe^{rt}[/tex]

[tex]50000=Pe^{0.08\times10}\\\\\Rightarrow\ 50000=P\times2.22554092849\\\\\Rightarrow\ P=\dfrac{50000}{2.22554092849}\\\\\Rightarrow\ P=22466.4482059\approx22,466.45[/tex]

Hence, the present value of asset would be $ 22,466.45.

The present value needed to obtain $50,000 in 10 years at an 8% continuously compounded interest rate is $22,466.48.

To determine the present value of an asset that grows to $50,000 in 10 years with an 8% annual compound interest rate, continuously compounded, we can use the formula for continuous compounding, which is:

A = Pe^rt

where:

A is the future value of the investment/loan, including interest,

P is the principal investment amount (the initial deposit or loan amount),

r is the annual interest rate (decimal),

t is the number of years the money is invested or borrowed for,

e is the base of the natural logarithm (approximately equal to 2.71828).

In this problem, we have A = $50,000, r = 0.08 (8% expressed as a decimal), and t = 10 years. We are solving for P, the present value.

Rearranging the formula to solve for P gives:

P = A / e^rt

P = 50000 / e^(0.08)(10)

Now calculate the value:

P = 50000 / e^0.8

P = 50000 / 2.22554... (using a calculator for e0.8)

P = $22,466.48 (rounded to two decimal places)

Thus, you would need to invest $22,466.48 now to have $50,000 in 10 years at an 8% annual compounded continuously interest rate.

Answer:

f(x) = 3x² - 12x  -135

Step-by-step explanation:

standard form of a quadratic equation is

y = Ax² + Bx + C

You are given 3 solutions for X and Y, i.e( x=-5, y = 0), (x = 9,y = 0) and  (x = 8,y = -39)

Substitute each of this equations into the quadratic equation  to obtain a system of 3 equations

For ( x=-5, y = 0), 25A - 5B + C = 0 ---------- eq (1)

For ( x= 9, y = 0), 81A + 9B + C = 0 ---------- eq (2)

For ( x= 8, y = -39), 64A + 8B + C = -39 ---------- eq (3)

You have 3 equations and 3 unknowns. Solving this system of 3 equations will give:

A = 3, B = -12, c = -135

Hence the quadratic equation is

y = 3x² - 12x -135

or in function form:

f(x) = 3x² - 12x  -135

Answer:

c

Step-by-step explanation:

i think not 100 percent sure

Answer:

CDI: 1.43/1.24x100= 115 What does this index mean? Good market potential.

Step-by-step explanation:

Answer: CDI: 1.43/1.24x100= 115 What does this index mean? Good market potential.

Step-by-step explanation:

Answer: Hence, the probability of having no girls is 0.0313.

Step-by-step explanation:

Since we have given that

Number of children a family has = 5

Number of outcomes would be [tex]2^5=32[/tex]

Probability of having a girl = [tex]\dfrac{1}{2}=0.5[/tex]

We need to find the probability of having no girls.

P(no girls ) = P( all boys )

So, it becomes,

[tex]P(all\ boys)=(0.5)^5=0.03125\approx 0.0313[/tex]

Hence, the probability of having no girls is 0.0313.

Answer:

  p(on schedule) ≈ 0.7755

Step-by-step explanation:

A suitable probability calculator can show you this answer.

_____

The z-values corresponding to the build time limits are ...

  z = (37.5 -45)/6.75 ≈ -1.1111

  z = (54 -45)/6.75 ≈ 1.3333

You can look these up in a suitable CDF table and find the difference between the values you find. That will be about ...

  0.90879 -0.13326 = 0.77553

The probability assembly will stay on schedule is about 78%.

Using the normal distribution, it is found that there is a 0.7747 = 77.47% probability that the build time will be such that assembly will stay on schedule.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

X = 54

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{54 - 45}{6.75}[/tex]

[tex]Z = 1.33[/tex]

[tex]Z = 1.33[/tex] has a p-value of 0.9082.

X = 37.5

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{37.5 - 45}{6.75}[/tex]

[tex]Z = -1.11[/tex]

[tex]Z = -1.11[/tex] has a p-value of 0.1335.

0.9082 - 0.1335 = 0.7747.

0.7747 = 77.47% probability that the build time will be such that assembly will stay on schedule.

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Answer:

A)0.5

Step-by-step explanation:

We can see in the graph , that it is bell-shaped along x =2. A bell-shaped graph along one value is called symmetric graph and it represents a normal distribution.

Since, the give graph is symmetric around x=2, so the mean of the data is 2.

The point immediate left to the mean represents x-σ

so,

2 - σ = 1.5

So,

σ = 0.5

The sigma represents standard deviation.

Hence, Option A is correct ..

Answer:

its A

Step-by-step explanation:

The simplification method would be the best to solve the given equation.

simplify means making it in a simple form by reducing variables in an equation.  we can achieve simplification easily by using PEMDAS.

Given equation 2x² - 7 = 9;

By simplify

2x² = 16

x² = 8

x = √8, -√8

Hence, for given equation simplification using PEMDAS is the best way of solving because it can be easily broken into parts to find the value of x.

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Answer: [tex]H_0:\mu=40[/tex]

[tex]H_0:\neq40[/tex]

Step-by-step explanation:

Claim 1. : Mean age of online dating service users is 40 years.

i.e. [tex]\mu=40[/tex], since it has equals sign so we take this as null hypothesis.

Claim 2. : Mean age of online dating service users is not 40 years.

[tex]\mu\neq40[/tex]

⇒ Null Hypothesis : [tex]H_0:\mu=40[/tex]

Alternative hypothesis : [tex]H_0:\neq40[/tex]

Answer:

D.

Step-by-step explanation:

Slope is rise over run by definition, and we are given the values for each in the problem.  The run is 80 and the rise is 15 so

[tex]m=\frac{15}{80}=\frac{3}{16}[/tex]

Answer:

120/s^6

Step-by-step explanation:

There is an easy formula for this...

L(t^n)=n!/(s^(n+1))

Your n=5 here

L(t^5)=5!/(s^6)

L(t^5)=120/s^6

[tex]\mathcal{L}\{t^n\}=\dfrac{n!}{s^{n+1}}[/tex]

So

[tex]\mathcal{L}\{f(t)\}=\dfrac{5!}{s^{5+1}}=\dfrac{120}{s^6}[/tex]

Answer:

true

Step-by-step explanation:

Answer:

True

Step-by-step explanation:

If 2 events are independent, then one event will not affect the other

Answer:

[tex]k(x)=2^{x}[/tex] ⇒ 1st answer

Step-by-step explanation:

* Lets explain how to solve the problem

∵ [tex]g(x)=3(2^{x})[/tex]

∵ [tex]h(x)=2^{x+1}[/tex]

- Lets revise this rule to use it

# If [tex]a^{n}*a^{m}=a^{n+m}====then==== a^{n+m}=a^{n}*a^{m}[/tex]

- We will use this rule in h(x)

∵ [tex]h(x)=2^{x+1}[/tex]

- Let a = 2 , n = x , m = 1

∴ [tex]h(x)=2^{x}*2^{1}[/tex]

- Now lets find k(x)

∵ k(x) = (g - h)(x)

∵ [tex]g(x)=3(2^{x})[/tex]

∵ [tex]h(x)=2^{x}*2^{1}[/tex]

∴ [tex]k(x)=3(2^{x})-(2^{x}*2^{1})[/tex]

- We have two terms with a common factor [tex]2^{x}[/tex]

∵ [tex]2^{x}[/tex] is a common factor

∵ [tex]\frac{3(2^{x})}{2^{x}}=3[/tex]

∵ [tex]\frac{2^{x}*2^{1}}{2^{x}}=2^{1}=2[/tex]

∴ [tex]k(x) = 2^{x}[3 - 2]=2^{x}(1)=2^{x}[/tex]

* [tex]k(x)=2^{x}[/tex]

Answer:

The equal balances will be $114 after 8 weeks

Step-by-step explanation:

* Lets study the information in the problem

- Manuel and Ruben both have bank accounts

- The system of equations models their balances y after x weeks

- Manuel balance is y = 11.5x + 22

- Ruben balance is y = -13x + 218

- After x weeks they will have same balances, means the values of y

 will be equal at the same values of x

- The solve the problem we will equate the two equations to find x

  and then substitute this x in on of the equation s to find the

  balance y

- Lets do that

∵ Manuel balance is y = 11.5x + 22

∵ Ruben balance is y = -13x + 218

∵ After x weeks their balances will be equal

- Equate the equations

∴ 11.5x + 22 = -13x + 218

- add 13 x for both sides

∴ 11.5x + 13x + 22 = 218

∴ 24.5x + 22 = 218

- subtract 22 from both sides

∴ 24.5x = 218 - 22

∴ 24.5x = 196

- Divide both sides by 24.5

∴ x = 8

- Their balances will be equals after 8 weeks

- To find the balance substitute x by 8 in any equation

∵ y = 11.5x + 22

∵ x = 8

∴ y = 11.5(8) + 22

∴ y = 92 + 22 = 114

∴ The equal balances will be $114

* The equal balances will be $114 after 8 weeks

Answer:

The equal balances will be $114 after 8 weeks

Step-by-step explanation:

Answer:

After how many years is the fish population 60?

x=2.71 years

Step-by-step explanation:

The fish population increases by a factor of 1.5 each year. We have the equation that represents this situation

[tex]f (x) = 20 (1.5) ^ x[/tex]

Where x represents the number of years elapsed f(x) represents the amount of fish.

Given this situation, the following question could be posed

After how many years is the fish population 60?

So we do [tex]f (x) = 60[/tex] and solve for the variable x

[tex]60 = 20 (1.5) ^ x\\\\\frac{60}{20} = (1.5)^x\\\\3 = (1.5)^x\\\\log_{1.5}(3) = log_{1.5}(1.5)^x\\\\log_{1.5}(3) = x\\\\x =log_{1.5}(3)\\\\x=2.71\ years[/tex]

Observe the solution in the attached graph

Answer: The answer should be D on edg

Please see the attachment for graph.

For x=1, Put x=1 into equation:

[tex]y = -4\sqrt{1} = - 4[/tex]

For x=4, Put x=4 into equation:

[tex]y = -4\sqrt{4} = - 8[/tex]

For x=9, Put x=9 into equation:

[tex]y = -4\sqrt{9} = - 12[/tex]

     1          -4

     4          -8

     9         -12

Now we plot the points on graph and join the points.

Please see the attachment for graph.

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The correct answer is -2 ln|x + 1| + 3 ln|x + 2| + C which corresponds to option (a).

We are asked to evaluate the integral:

∫ (x - 1) ÷ (x² + 3x + 2) dx

First, factor the denominator:

x² + 3x + 2 = (x + 1)(x + 2)

This allows us to use partial fraction decomposition to rewrite the integral :

(x - 1) ÷ [(x + 1)(x + 2)] = A ÷ (x + 1) + B ÷ (x + 2)

Next, solve for A and B:

So, we can write :

(x - 1) ÷ [(x + 1)(x + 2)] = -2 ÷ (x + 1) + 3 ÷ (x + 2)

Integrate both terms separately :

∫ (-2 ÷ (x + 1)) dx + ∫ (3 ÷ (x + 2)) dx

This gives us :

-2 ln|x + 1| + 3 ln|x + 2| + C

Hence, the solution is :

-2 ln|x + 1| + 3 ln|x + 2| + C

The correct answer is option (a).

Complete Question :

Use Partial fraction expansion to evaluate : ∫ (x - 1) ÷ (x² + 3x + 2) dx

a. -2 ln|x + 1| + 3 ln|x + 2| + C   b. [tex]\frac{-2}{x+1}+\frac{3}{x+2}+C - 2 x + 1 + 3 x + 2 + C[/tex]

c. [tex]\frac{2}{\left(x+1\right)^2}+\frac{-3}{\left(x+2\right)^2}+C 2 ( x + 1 ) 2 + - 3 ( x + 2 ) 2 + C[/tex]

d. [tex]\frac{1}{\left(x+3+\frac{2}{x}\right)^2}+C 1 ( x + 3 + 2 x ) 2 + C[/tex]

Answer:

13.4%

Step-by-step explanation:

Use binomial probability:

P = nCr p^r q^(n-r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

Here, n = 16, r = 2, p = 0.25, and q = 0.75.

P = ₁₆C₂ (0.25)² (0.75)¹⁶⁻²

P = 120 (0.25)² (0.75)¹⁴

P = 0.134

There is a 13.4% probability that exactly 2 students will withdraw.

The probability that exactly 2 out of 16 students will withdraw from an introductory statistics class, given a historical withdrawal rate of 25%, can be calculated using the binomial probability formula.

This problem falls into the category of binomial probability. We define 'success' as a student withdrawing from the course. The number of experiments is 16 (as there are 16 students), the number of successful experiments we are interested in is 2 (we want to know the probability of exactly 2 student withdrawing), and the probability of success on a single experiment is 0.25 (as per the given 25% withdrawal rate).

To calculate binomial probability, we can use the binomial formula P(X=k) = C(n, k)*(p^k)*((1-p)^(n-k)), where:
P(X=k) = probability of k successes
C(n, k) = combination of n elements taken k at a time
p = probability of success
n, k = number of experiments, desired number of successes respectively.

Substituting our values into this formula, we get:
P(X=2) = C(16, 2) * (0.25^2) * ((1-0.25)^(16-2)).

You will have to calculate the combination and simplify the expression to get your final probability.

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Answer:

The last choice is the one you want

Step-by-step explanation:

If you plug in the values of x to our rational function, the y values you get back match those in the last choice.  The limit is -4; we see that as our x value approach 0 (but cannot equal 0!!), the y values get closer and closer to -4.  So that's the limit!

Answer:

[tex]\large\boxed{(-7x+4)(-7x-4)}[/tex]

Step-by-step explanation:

[tex]\text{The difference of squares:}\ a^2-b^2=(a-b)(a+b)\\\\(-7x+4)(-7x-4)=(-7x)^2-4^2=49x^2-16[/tex]

(-7x + 4) (-7x - 4) can be written as a difference of squares.

Option C is the correct answer.

An expression contains one or more terms with addition, subtraction, multiplication, and division.

We always combine the like terms in an expression when we simplify.

We also keep all the like terms on one side of the expression if we are dealing with two sides of an expression.

Example:

1 + 3x + 4y = 7 is an expression.com

3 + 4 is an expression.

2 x 4 + 6 x 7 – 9 is an expression.

33 + 77 – 88 is an expression.

We have,

The difference of squares is a special algebraic form that occurs when we multiply two binomials of the form (a + b)(a - b).

This results in the product of two terms:

The square of the first term minus the square of the second term.

In other words, we have (a + b)(a - b) = a² - b².

In the given options, only (-7x + 4) (-7x - 4) can be written as a difference of squares, by applying the above formula.

We can rewrite it as:

(-7x + 4) (-7x - 4) = (-7x)² - 4² = 49x² - 16

The other options do not follow this pattern and cannot be written as a difference of squares.

Thus,

(-7x + 4)(-7x - 4) = 49x² - 16

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Answer:

573 L is im sure the correct answer

Answer with explanation:

1.

Let a, and b be two numbers between 20 and 5 , which is in geometric progression.

So,the series is as Follows =20 , a, b, 5

Common ratio

          [tex]=\frac{\text{Second term}}{\text{First term}}[/tex]

[tex]\frac{20}{a}=\frac{a}{b}=\frac{b}{5}\\\\b^2=5 a---(1)\\\\a^2=20 b\\\\\frac{b^4}{25}=20 b-----\text{Using 1}\\\\b^3=500\\\\b=(500)^{\frac{1}{3}}\\\\b=5\times (4)^{\frac{1}{3}}\\\\5a=25\times (4)^{\frac{2}{3}}\\\\a=5\times (4)^{\frac{2}{3}}[/tex]

2.

44 -32-3

=12-3

=9

3.

⇒Sin (2.4)=Sin(2+0.4)

⇒Sin 2 ×Cos (0.4)+Cos 2 × Sin (0.4)

⇒Sin (A+B)=Sin A×Cos B+Cos A×Sin B

Answer with explanation:

Let, A=[1,1,2]

B=[1,2,1]

C=[2,1,5]

⇒Now, Writing vector , A in terms of Linear combination of C and B

A=x B +y C

⇒[1,1,2]=x× [1,2,1] + y×[2,1,5]

1.→1 = x +2 y

2.→ 1=2 x +y

3.→ 2= x+ 5 y

Equation 3 - Equation 1

→3 y=1

[tex]y=\frac{1}{3}[/tex]

[tex]1=x+\frac{2}{3}\\\\x=1 -\frac{2}{3}\\\\x=\frac{1}{3}[/tex]

So, Vector A , can be written as Linear Combination of B and C.

⇒Now, Writing vector , B in terms of Linear combination of A and C

Now, let, B = p A+q C

→[1,2,1]=p× [1,1,2] +q ×[2,1,5]

4.→1= p +2 q

5.→2=p +q

6.→1=2 p +5 q

Equation 5 - Equation 4

-q =1

q= -1

→2= p -1

→p=2+1

→p=3

So, Vector B , can be written as Linear Combination of A and C.

⇒Now, Writing vector , C in terms of Linear combination of A and B

C=m A + n B

[2,1,5] = m×[1,1,2] + n× [1,2,1]

7.→2= m+n

8.→1=m +2 n

9.→5=2 m +  n

Equation 8 - Equation 7

n= -1

→m+ (-1)=2

→m=2+1

→m=3

So, Vector C , can be written as Linear Combination of A and B.

So, All the three vectors , A=[1,1,2],B=[1,2,1],C=[2,1,5] can be written as Linear combination of each other.

⇒≡But , the two vectors, (0.4,3.7,-1.5) (0.2,0),can't be written as Linear combination of each other as first vector is of order, 1×3, and second is of order, 1×2.

None of the selections is correct.

In this case, we must check the existence of a set of real coefficients such that the following two linear combinations exist:

[tex]\alpha_{1}\cdot (1, 1, 2)+\alpha_{2}\cdot (1, 2,1)+\alpha_{3}\cdot (2, 1, 5) = (0.4, 3.7, -1.5)[/tex]   (1)

[tex]\alpha_{4}\cdot (1,1,2)+\alpha_{5}\cdot (1,2,1) + \alpha_{6}\cdot (2,1,5) = (0, 2, 0)[/tex]   (2)

Now we proceed to solve each linear combination:

[tex]\alpha_{1}+\alpha_{2}+2\cdot \alpha_{3} = 0.4[/tex]

[tex]\alpha_{1}+2\cdot \alpha_{2}+\alpha_{3} = 3.7[/tex]

[tex]2\cdot \alpha_{1}+\alpha_{2}+5\cdot \alpha_{3} = -1.5[/tex]

The system has no solution, since the third equation is a linear combination of the first and second ones.

[tex]\alpha_{4}+\alpha_{5}+2\cdot \alpha_{6} = 0[/tex]

[tex]\alpha_{4}+2\cdot \alpha_{5}+\alpha_{6} = 2[/tex]

[tex]2\cdot \alpha_{4}+\alpha_{5}+5\cdot \alpha_{6} = 0[/tex]

The system has no solution, since the third equation is a linear combination of the first and second ones.

None of the selections is correct. [tex]\blacksquare[/tex]

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Answer:

1. A: 0.25; B: 0.03; C: 1.41; D: -0.28

2. A: 0.39; B: 0.06; C: 40.30; D: 21.81

Step-by-step explanation:

For CDF lookups, we used the Excel NORMDIST(x, mean, stdev, TRUE) function. For inverse CDF lookups, we used the NORMINV(x, mean, stdev) function.

Each of these functions works with the area under the curve from -∞ to x, so for cases where we're interested in the upper tail, we subtract the probability from 1, or subtract the x value from twice the mean.

For question 1, we computed the Z values in each case. The NORMDIST function works directly with x, mean, and standard deviation, so does not need the z value.

Answer:

(a) [tex]f'(x)=-\frac{2}{x^3}[/tex]

(b) [tex]y=-0.25x+0.75[/tex]

Step-by-step explanation:

The given function is

[tex]f(x)=\frac{1}{x^2}[/tex]                  .... (1)

According to the first principle of the derivative,

[tex]f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}[/tex]

Cancel out common factors.

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}[/tex]

By applying limit, we get

[tex]f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}[/tex]

[tex]f'(x)=\frac{-2x)}{x^4}[/tex]

[tex]f'(x)=\frac{-2)}{x^3}[/tex]                         .... (2)

Therefore [tex]f'(x)=-\frac{2}{x^3}[/tex].

(b)

Put x=2, to find the y-coordinate of point of tangency.

[tex]f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25[/tex]

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

[tex]m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}[/tex]

Substitute x=2 in equation 2.

[tex]f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25[/tex]

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

[tex]y-y_1=m(x-x_1)[/tex]

[tex]y-0.25=-0.25(x-2)[/tex]

[tex]y-0.25=-0.25x+0.5[/tex]

[tex]y=-0.25x+0.5+0.25[/tex]

[tex]y=-0.25x+0.75[/tex]

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

Answer:

12 a. 4605 feet 12 b. 1,459,063 square feet

Step-by-step explanation:

For the perimeter, we simply add the lengths of each of the 5 sides together (or multiply 5 times one side length).

P = 5(921)

P = 4605 feet

For the area, we will use composition...add the area of the triangle to the area of the trapezoid.

For the area of the triangle, the formula is

[tex]A=\frac{1}{2}bh[/tex].

Filling in our values gives us

[tex]A=\frac{1}{2}(1490)(541)[/tex] and

A = 403,045 square feet.

Now for the trapezoid.  The formula for a trapezoid is

[tex]A=\frac{1}{2}(b_{1}+b_{2})(h)[/tex]

where the b's represent the bases and the h represents the height.  Filling in our values gives us

[tex]A=\frac{1}{2}(921+1490)(876)[/tex]

Work inside the parenthesis first:

[tex]A=\frac{1}{2}(2411)(876)[/tex] and

A = 1,056,018

Now we add those together to get that area of the Pentagon is 1,459,063 square feet

Answer: a.- discrete   b.- continous

Step-by-step explanation: Discrete Variable. Variables that can only take on a finite number of values are called "discrete variables." All qualitative variables are discrete. Some quantitative variables are discrete, such as performance rated as 1,2,3,4, or 5, or temperature rounded to the nearest degree.

Continuous Variable. If a variable can take on any value between its minimum value and its maximum value, it is called a continuous variable; otherwise, it is called a discrete variable.

The number of fish caught during a fishing tournament is a discrete variable, and the time it takes for a light bulb to burn out is a continuous variable

Let's define what variables are. Variables are any representation of a phenomenon or property that changes over time. In simple terms, variables are "things" that change, meaning they don't have a constant value. Variables can be either discrete or continuous.

To understand these concepts it's better to understand first what continuous means, continuous variables are those which can take any value whatsoever over time. This last statement is the main idea but it's not self-explanatory, a test to check whether a variable is continuous or not is to take any 2 possible outcomes of that variable, and check if that variable can take any value between those 2 possible outcomes. If the test gives positive then our variable is continuous, if not then it's discrete.

Let's test the first question. During a fishing tournament, each person can fish only one fish at a time, therefor possible outcomes are 1 fish, or 2 fish, or 3, or 4, and so on. This means that we will never be able to get, for example, 1.5 fish (which is a value between 2 possible outcomes, 1 fish and 2 fish), therefor our variable is discrete.

Let's test the second question. The time it takes for a light bulb to burn out has many possible outcomes, examples are 1 second, 2.5 seconds, 10 minutes, etc. If we check between any of those possible outcomes, we will always be able to find a time, doesn't matter how precise, in which the light bulb could burn. This means that the time for a light bulb to burn is continuous.

Variable, Continuous, Discrete, Interval

Answer:

1) C. 2

2) A. 2

Step-by-step explanation:

1. We need to descompose 256 into its prime factors:

[tex]256=2*2*2*2*2*2*2*2=2^8[/tex]

We must rewrite the expression [tex]\sqrt[8]{256}[/tex]:

[tex]=\sqrt[8]{2^8}[/tex]

We need to remember that:

[tex]\sqrt[n]{a^n}=a[/tex]

Then:

[tex]=2[/tex]

2. Let's descompose 16 into its prime factors:

[tex]16=2*2*2*2=2^4[/tex]

We must rewrite the expression [tex]\sqrt[4]{16}[/tex]:

[tex]=\sqrt[4]{2^4}[/tex]

Then we get:

[tex]=2[/tex]

For [tex]\(\sqrt[8]{256}\)[/tex] , the answer is 2 (C), and for [tex]\(\sqrt[4]{16}\)[/tex], the answer is also 2(A), obtained through prime factorization and simplifying using the property [tex]\(\sqrt[n]{a^n} = a\)[/tex].

Let's go into more detail for both questions:

1. [tex]\(\sqrt[8]{256}\)[/tex]:

  - First, find the prime factorization of 256: [tex]\(256 = 2^8\)[/tex]

  - Rewrite the expression as [tex]\(\sqrt[8]{2^8}\)[/tex].

  - Using the property [tex]\(\sqrt[n]{a^n} = a\)[/tex], simplify to 2.

  - Therefore, [tex]\(\sqrt[8]{256} = 2\)[/tex]

  - Correct answer: C. 2

2. [tex]\(\sqrt[4]{16}\)[/tex]:

  - Start with the prime factorization of 16: [tex]\(16 = 2^4\)[/tex]

  - Express the expression as [tex]\(\sqrt[4]{2^4}\)[/tex]

  - Apply the property  [tex]\(\sqrt[n]{a^n} = a\) to get \(2\).[/tex]

  - Thus,  [tex]\(\sqrt[4]{16} = 2\)[/tex]

  - Correct answer: A. 2

In both cases, understanding the prime factorization and utilizing the property of radicals [tex](\(\sqrt[n]{a^n} = a\))[/tex] helps simplify the expressions and find the correct values.

Answer: 0.9826

Step-by-step explanation:

Given : Mean : [tex]\lambda =3\text{ calls per minute}[/tex]

For two minutes period the new mean would be :

[tex]\lambda_1=2\times3=6\text{ calls per two minutes}[/tex]

The formula to calculate the Poisson distribution is given by :_

[tex]P(X=x)=\dfrac{e^{-\lambda_1}\lambda_1^x}{x!}[/tex]

Then ,the required probability is given by :-

[tex]P(X\geq2)=1-(P(X\leq1))\\\\=1-(P(0)+P(1))\\\\=1-(\dfrac{e^{-6}6^0}{0!}+\dfrac{e^{-6}6^1}{1!})\\\\=1-0.0173512652367\\\\=0.982648734763\approx0.9826[/tex]

Hence, the probability that at least two calls are received in a given two-minute period is 0.9826.

Answer:

  1287

Step-by-step explanation:

The number of combinations of 13 diamonds taken 5 at a time is ...

  13C5 = 13·12·11·10·9/(5·4·3·2·1) = 13·11·9 = 1287