Determine whether the random variable is discrete or continuous. In each​ case, state the possible values of the random variable. ​(a) The number of fish caught during a fishing tournament . ​(b) The time it takes for a light bulb to burn out .

Answer: The mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 is

[tex]113.9\text{ and }4.0[/tex] respectively.

Step-by-step explanation:

Given : The mean of sampled population : [tex]\mu = 113.9[/tex]

The standard deviation of sampled population : [tex]\sigma = 32.1[/tex]

We know that the mean and standard deviation of the sampling distribution of the sample mean for samples of size n is given by :_

[tex]\mu_s=\mu\\\\\sigma_s=\dfrac{\sigma}{\sqrt{n}}[/tex]

Now, the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 will be :-

[tex]\mu_s=113.9\\\\\sigma_s=\dfrac{32.1}{\sqrt{64}}=4.0125\approx4.0[/tex]

The mean of the sample is 113.9, and the standard deviation of the sample is 4.0

The given parameters are:

[tex]\mu = 113.9[/tex] --- the population mean

[tex]\sigma = 32.1[/tex] --- the population standard deviation

[tex]n = 64[/tex] --- the sample size

The sample mean is calculated as:

[tex]\bar x = \mu[/tex]

So, we have:

[tex]\bar x = 113.9[/tex]

The sample standard deviation is calculated as:

[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]

This gives

[tex]\sigma_x = \frac{32.1}{\sqrt {64}}[/tex]

[tex]\sigma_x \approx 4.0[/tex]

Hence, the mean of the sample is 113.9, and the standard deviation of the sample is 4.0

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Explanation:

Assuming the side of the boat is a straight line, it would constitute a transversal crossing the lines of the oars. When corresponding angles at a transversal are congruent, the lines being crossed are parallel. Since the oars are those lines, the oars are parallel.

Answer:

Step-by-step explanation:

so u do 2+2 4=243==32===3=424=4=234=234=32=43=4=34

Answer:

  If x is a multiple of 4, then x/2 is even.

Step-by-step explanation:

An integer is even, if it is equal to 2n for some integer n. We want x/2 to be even, so ...

  x/2 = 2n

  x = 4n . . . . . multiply by 2

That is, x will be equal to 4n for some integer n. x is a multiple of 4.

To determine a sufficient condition for x/2 to be an even integer, we need to find values of x for which x/2 results in an even integer.

To find a sufficient condition for x/2 to be an even integer, we need to determine for which values of x the expression x/2 results in an even integer. Since an even integer is divisible by 2 without a remainder, a sufficient condition for x/2 to be an even integer is that x itself is divisible by 2 without a remainder. In other words, x should be an even integer.

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Answer:

There are two times for the ball to reach a height of 64 feet:

1 second after thrown ⇒ the ball moves upward

2 seconds after thrown ⇒ the ball moves downward

Step-by-step explanation:

* Lets explain the function to solve the problem

- h(t) models the height of the ball above the ground as a function

 of the time t

- h(t) = -16t² + 48t + 32

- Where h(t) is the height of the ball from the ground after t seconds

- The ball is thrown upward with initial velocity 48 feet/second

- The ball is thrown from height 32 feet above the ground

- The acceleration of the gravity is -32 feet/sec²

- To find the time when the height of the ball is above the ground

  by 64 feet substitute h by 64

∵ h(t) = -16t² + 48t + 32

∵ h = 64

∴ 64 = -16t² + 48t + 32 ⇒ subtract 64 from both sides

∴ 0 = -16t² + 48t - 32 ⇒ multiply the both sides by -1

∴ 16t² - 48t + 32 = 0 ⇒ divide both sides by 16 because all terms have

  16 as a common factor

∴ t² - 3t + 2 = 0 ⇒ factorize it

∴ (t - 2)(t - 1) = 0

- Equate each bracket by zero to find t

∴ t - 2 = 0 ⇒ add 2 to both sides

∴ t = 2

- OR

∴ t - 1 = 0 ⇒ add 1 to both sides

∴ t = 1

- That means the ball will be at height 64 feet after 1 second when it

 moves up and again at height 64 feet after 2 seconds when it

 moves down

* There are two times for the ball to reach a height of 64 feet

  1 second after thrown ⇒ the ball moves upward

  2 seconds after thrown ⇒ the ball moves downward

Answer:

the equation is true only if c=6 and d=2.

Step-by-step explanation:

We have the following expression:

[tex]\sqrt[3]{162x^{c}y^{5}} = 3x^{2}y\sqrt[3]{6y^{d}}[/tex]

Elevating to the power of three:

[tex]162x^{c}y^{5}=27x^{6}y^{3}(6y^{d})[/tex]

Simplifying:

→ [tex]162x^{c}y^{5}=162x^{6}y^{3}y^{d}[/tex]

→ [tex]x^{c}y^{5}=x^{6}y^{3}y^{d}[/tex]

→ [tex]x^{c}y^{5}=x^{6}y^{d+3}[/tex]

By comparing the two expression, we can say that:

[tex]c=6[/tex]

[tex]d+3 = 5[/tex] → [tex]d=2[/tex]

Therefore, the equation is true only if c=6 and d=2.

Answer:

c = 6 d = 2

on edge

Answer:

=16.492

Step-by-step explanation:

3√17 is a surd that can be broken into 3 × √17

√17 = 4.123 ( to 4 S.F )

3 × 4.123 = 16.492

The decimal form is an estimation as it has many digits that can only be rounded off or truncated.

Answer:

10.6%

Step-by-step explanation:

Normal curves are symmetrical.  That means that on a standard normal distribution,  the area less than -1.25 is the same as the area greater than +1.25.  The total area under the curve is 1, so:

P = 1 - 0.894

P = 0.106

Approximately 10.6% of the area under the curve lies below -1.25.

For a standard normal distribution, the area under the curve to the left of a z-score of -1.25 is approximately 10.6% due to the fact that a normal distribution is symmetric around its mean.

In a standard normal distribution, the properties of symmetry mean that the area on either side of the mean (µ=0) is identical. When looking at positive and negative z-scores that are the same magnitude but opposite in direction, the areas under the curve to their respective sides will be equivalent.

The given z-score is 1.25 and we know the area under the curve to the left of this z-score is 0.894 or 89.4%. Because of symmetry, the z-score of -1.25 will have an equal area under the curve to the right, which also represents 89.4%. Therefore, the area under the curve to the left of a z-score of -1.25 is 1 - 0.894 = 0.106, or approximately 10.6%.

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Answer:

The solution is [tex]x\leq 1[/tex]

All real numbers less than or equal to 1

The graph in the attached figure

Step-by-step explanation:

we have

[tex]8\geq 3x+5[/tex]

Subtract 5 both sides

[tex]8-5\geq 3x[/tex]

[tex]3\geq 3x[/tex]

Divide by 3 both sides

[tex]1\geq x[/tex]

Rewrite

[tex]x\leq 1[/tex]

The solution is the interval ------> (-∞,1]

All real numbers less than or equal to 1

In a number line the solution is the shaded area at left of x=1 (close circle)

see the attached figure

Try this solution:

There are several ways to find the max or min of the given function:

1. to use derivative of the function. For more details see the attachment (3 basic steps); the coordinates of max-point are marked with green (-5; 14.5)

2. to use formulas. The given function is the standart function with common equation y=ax²+bx+c, it means the correspond formulas are (where a<0, the vertex of this function is its maximum):

[tex]X_0=-\frac{b}{2a} ; \ X_0=-\frac{-5}{2*(-\frac{1}{2})} =-5.[/tex]

[tex]Y_0=-\frac{D}{4a}; \ Y_0=-\frac{25+4*2*0.5}{4*(-\frac{1}{2})} =14.5[/tex]

Finally: point (-5;14.5) -  maximum of the given function.

3. to draw a graph.

Answer:

The largest area is 125000 m²

The dimensions of the farmland are 250 m and 500 m

Step-by-step explanation:

* Lets pick the information from the problem

- The farmland is shaped a rectangle

- The farmland will be bounded on one side by a river

- The other three sides are bounded by a​ single-strand electric fence

- The length of wire is 1000 m

- Lets consider the width of the rectangle is x and the length is y

- The side which will be bounded by the river is y

∴ The perimeter of the farmland which will be bounded by the electric

  fence = x + x + y  = 2x + y

- We will use the wire to fence the farmland

∵ The length of the wire is 1000 m

∵ The perimeter of the farmland is equal to the length of the wire

∴ 2x + y = 1000

- Lets find y in term of x

∵ 2x + y = 1000 ⇒ subtract 2x from both sides

∴ y = 1000 - 2x

- Now lets find the area can enclose by the wire

∵ The area of the rectangle = length × width

∵ The width of the farmland is x and its length is y

∴ The area of the farmland (A) = x × y = xy ⇒ (2)

- Use equation (1) to substitute the value of y in equation (2)

∴ A = x (1000 - 2x) ⇒ simplify

∴ A = 1000 x - 2 x²

- To find the maximum area we will differentiate A with respect to x

  and equate the answer by zero to find the value of x which will make

  the enclosed area largest

* Lets revise the rule of differentiation

- If y = ax^n, then dy/dx = a(n) x^(n-1)

- If y = ax, then dy/dx = a

- If y = a, then dy/dx = 0 , where a is a constant

∵ A = 1000 x - 2 x² ⇒ (3)

- Differentiate A with respect to x using the rules above

∴ dA/dx = 1000 - 2(2) x^(2-1)

∴ dA/dx = 1000 - 4x

- Put dA/dx = 0 to find the value of x

∵ 1000 - 4x = 0 ⇒ add 4x to both sides

∴ 1000 = 4x ⇒ divide both sides by 4

∴ 250 = x

∴ The value of x is 250

- Lets substitute this value in equation 3 to find the largest area

∵ A = 1000 x - 2 x²

∴ A = 1000 (250) - 2(250)² = 125000 m²

* The largest area is 125000 m²

∵ The width of the farmland is x

∵ x = 250

∴ The width of the farmland = 250 m

- Substitute the value of x in the equation (1) to find y

∵ y = 1000 - 2x

∵ x = 250

∴ y = 1000 - 2(250) = 1000 - 500 = 500

∵ The length of the farm lend is y

∴ The length of the farm land = 500 m

* The dimensions of the farmland are 250 m and 500 m

To get the maximum area from a rectangular farmland bounded on one side by a river and on the other three sides by a single-strand electric fence with 1000 meters of wire, the dimensions should be 250m x 500m yielding a maximum area of 125,000 square meters.

This problem can be approached as a classic calculus maximization problem. The scenario mentioned in your question is about maximizing the area of a rectangle with a constant perimeter, this occurs when the rectangle is a square.

The three sides of your plot will consume the 1000 m of wire, if the length of each adjacent sides are x and y (where x is the length of the fence along the river, and y is the length of the other two fences), then

y + 2x = 1000

Substituting y from this equation into the area equation, A = xy(which results in A = x(1000-2x)), you can differentiate this equation with regards to x and set the derivative equal to zero to find the x-value that will give the maximum area. This happens when x = 250m,  y = 500m then the maximum area is 125,000 m².

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Answer:

Step-by-step explanation:

You know that fractions with the same value in numerator and denominator reduce to 1. This is true whether the value is a number, a variable expression, or some mix of those. That is ...

[tex]\dfrac{1760}{1760}=1\\\\\dfrac{3\,\text{mi}}{1\,\text{mi}}=\dfrac{3}{1}\cdot\dfrac{\text{mi}}{\text{mi}}=3[/tex]

This example should show you that you can treat units as if they were a variable.

So, the unit conversion process is the process of choosing combinations of numerator and denominator units so that all the units you don't want cancel, leaving only units you do want.

You're starting with a number than has "laps" in the numerator. To cancel that, you need to find a conversion factor with "lap" in the denominator. On the list you are given, the one that has that is ...

  [tex]\dfrac{1920\,\text{yd}}{1\,\text{lap}}[/tex]

Now, you have canceled laps, but you have yards. Also on your list of conversion factors is a ratio with yards in the denominator:

  [tex]\dfrac{3\,\text{ft}}{1\,\text{yd}}[/tex]

This will cancel the yards in the numerator from the previous result, but will give you feet in the numerator. You want miles, so you look for a conversion factor between feet and miles, with miles in the numerator. The one you find is ...

  [tex]\dfrac{1\,\text{mi}}{5280\,\text{ft}}[/tex]

These three conversion factors go into the blanks. When you form the product, you will get ...

  [tex]\dfrac{38.5\cdot 1920\cdot 3}{1\cdot 1\cdot 5280}\cdot\dfrac{\text{laps$\cdot$yd$\cdot$ft$\cdot$mi}}{\text{lap$\cdot$yd$\cdot$ft}}=42\,\text{mi}[/tex]

Answer:

Step-by-step explanation:

The value of t is:

                     t=48.8889

We are asked to solve the linear equation in terms of variable t.

The equation is given by:

[tex]0.95t+0.05(100-t)=0.49(100)[/tex]

Firstly we will solve the parentheses term in the left and right hand side of the equality as follows:

[tex]0.95t+0.05\times 100-0.05t=49[/tex]

Now we combine the like terms on the left side of equality by:

[tex]0.95t-0.05t+5=49\\\\i.e.\\\\0.90t+5=49[/tex]

Now we subtract both side of the equation by 5 to get:

[tex]0.90t=44[/tex]

Now on dividing both side of the equation by 0.90 we get:

[tex]t=\dfrac{44}{0.90}\\\\i.e.\\\\t=48.8889[/tex]

Hence, the value of t is:   48.8889

After simplifying and rearranging the given equation, the solution for the variable 't' comes out to be approximately 48.89. The solution was verified by substituting 't' back into the original equation.

The equation given is: 0.95t + 0.05(100 - t) = 0.49(100). Start by simplifying the left side of the equation, which gives us: 0.95t + 5 - 0.05t = 49. Combine the t terms to get 0.9t + 5 = 49. Rearranging for t gives us 0.9t = 44. Dividing both sides by 0.9 yields t = 48.89. Checking our answer, we can substitute t back into the original equation: 0.95(48.89) + 0.05(100 - 48.89) = 49; simplifying this, we get 46.45 + 2.56 = 49, which checks out.

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Answer:

[tex] log ( \frac { 5 x } { 4 y} ) \implies [/tex] [tex] log ( 5 ) + log ( x ) - log ( 4 ) + log ( y ) [/tex]

Step-by-step explanation:

We are given the following for which we are to expand the logarithm:

[tex] log ( \frac { 5 x } { 4 y} ) [/tex]

Expanding the log by applying the rules of expanding the logarithms by changing the division into subtraction:

[tex] log ( 5 x ) - log ( 4 y ) [/tex]

[tex] log ( 5 ) + log ( x ) - log ( 4 ) + log ( y ) [/tex]

[tex]\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log\left( \cfrac{5x}{4y} \right)\implies \log(5x)-\log(4y)\implies [\log(5)+\log(x)]-[\log(4)+\log(y)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \log(5)+\log(x)-\log(4)-\log(y)~\hfill[/tex]

First, lets convert them into multipliers:

The multiplier for a:

                             25% increase = 1.25

                             40% decrease = 0.6

                             40% increase  = 1.4

Now to work out the overall percentage change, we just times all of the multipliers together, and convert it back to a percentage:

1.25 x 0.6 x 1.4 = 1.05    

So the overall multiplier is 1.05

And a multiplier of 1.05 = a 5% increase.

That means that the percentage change is + 5%

_________________________________________

Answer:

The percentage change in shares from the start of 2013 to the end of 2015 is:

+ 5%

_______________________________________

Note: if you haven't been taught multipliers - then ask and I'll try my best to explain!

Answer:

B. 1.25

Step-by-step explanation:

Probability is as below

[tex]0 \leqslant p(a) \leqslant 1[/tex]

When P(A) = 0, it is an unlikely event

When P(A) = 1, it is a certain event

Answer:

450 km

Step-by-step explanation:

Let's say Va is the speed of the car from city A, Ta is the time it spent traveling, and Da is the distance it traveled.

Similarly, Vb is the speed of the car from city B, Tb is the time it spent traveling, and Db is the distance it traveled.

Given:

Va = Vb - 10

Ta₁ = Tb₁ = 5

Ta₂ = Tb₂ + 4.5

Db₂ = 150

Find:

D = Da₁ + Db₁ = Da₂ + Db₂

Distance = rate × time

In the first scenario:

Da₁ = Va Ta₁

Da₁ = (Vb - 10) (5)

Da₁ = 5Vb - 50

Db₁ = Vb Tb₁

Db₁ = Vb (5)

Db₁ = 5Vb

So:

D = Da₁ + Db₁

D = 10Vb - 50

In the second scenario:

Da₂ = Va Ta₂

Da₂ = (Vb - 10) (Tb₂ + 4.5)

Da₂ = Vb Tb₂ + 4.5Vb - 10Tb₂ - 45

Db₂ = Vb Tb₂

150 = Vb Tb₂

Substituting:

Da₂ = 150 + 4.5Vb - 10Tb₂ - 45

Da₂ = 105 + 4.5Vb - 10Tb₂

Da₂ = 105 + 4.5Vb - 10 (150 / Vb)

Da₂ = 105 + 4.5Vb - (1500 / Vb)

So:

D = Da₂ + Db₂

D = 105 + 4.5Vb - (1500 / Vb) + 150

D = 255 + 4.5Vb - (1500 / Vb)

Setting this equal to the equation we found for D from the first scenario:

10Vb - 50 = 255 + 4.5Vb - (1500 / Vb)

5.5Vb - 305 = -1500 / Vb

5.5Vb² - 305Vb = -1500

5.5Vb² - 305Vb + 1500 = 0

11Vb² - 610Vb + 3000 = 0

(Vb - 50) (11Vb - 60) = 0

Vb = 50, 5.45

Since Vb > 10, Vb = 50 km/hr.

So the distance between the cities is:

D = 10Vb - 50

D = 10(50) - 50

D = 450 km

Let's denote the events as follows:

- A: Module A fails

- B: Module B fails

Given probabilities:

[tex]\[ P(A) = 0.24 \][/tex]

[tex]\[ P(B) = 0.38 \][/tex]

We can use the addition rule of probability for mutually exclusive events to calculate the probabilities:

(a) Probability that only Module A fails:

[tex]\[ P(A \text{ only}) = P(A) = 0.24 \][/tex]

(b) Probability that only Module B fails:

[tex]\[ P(B \text{ only}) = P(B) = 0.38 \][/tex]

(c) Probability that both modules fail:

[tex]\[ P(\text{both}) = P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]

(d) Probability that neither module fails:

[tex]\[ P(\text{neither}) = 1 - (P(A) + P(B) - P(A \cap B)) \][/tex]

[tex]\[ = 1 - (0.24 + 0.38 - 0.0912) \][/tex]

[tex]\[ = 1 - 0.5288 \][/tex]

[tex]\[ = 0.4712 \][/tex]

(e) The probabilities should add up to 1, representing the total probability of all possible outcomes:

[tex]\[ P(A \text{ only}) + P(B \text{ only}) + P(\text{both}) + P(\text{neither}) = 0.24 + 0.38 + 0.0912 + 0.4712 = 1 \][/tex]

So, the probabilities add up to 1, as expected.

a. The probability that only Module A fails is 0.1488

b. The probability that only Module B fails is 0.2888

c. The probability that both modules fail is 0.0912

d. The probability that neither module fails is 0.4712

e. The probabilities add up to is 1

Let's denote the events as follows:

Event A: Module A fails

Event B: Module B fails

Given probabilities:

P(A) = 0.24

P(B) = 0.38

(a) To find the probability that only Module A fails:

[tex]\[ P(A \cap \neg B) = P(A) \times P(\neg B) \][/tex]

To calculate [tex]\( P(\neg B) \)[/tex], we subtract the probability of B from 1:

[tex]\[ P(\neg B) = 1 - P(B) = 1 - 0.38 = 0.62 \][/tex]

[tex]\[ P(A \cap \neg B) = 0.24 \times 0.62 = 0.1488 \][/tex]

(b) Similarly, to find the probability that only Module B fails:

[tex]\[ P(\neg A \cap B) = P(\neg A) \times P(B) \][/tex]

[tex]\[ P(\neg A) = 1 - P(A) = 1 - 0.24 = 0.76 \][/tex]

[tex]\[ P(\neg A \cap B) = 0.76 \times 0.38 = 0.2888 \][/tex]

(c) To find the probability that both modules fail:

[tex]\[ P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]

(d) To find the probability that neither module fails:

[tex]\[ P(\neg A \cap \neg B) = P(\neg A) \times P(\neg B) \][/tex]

[tex]\[ P(\neg A \cap \neg B) = 0.76 \times 0.62 = 0.4712 \][/tex]

(e) The probabilities add up to :

[tex]\[ P(A \cap \neg B) + P(\neg A \cap B) + P(A \cap B) + P(\neg A \cap \neg B) = 0.1488 + 0.2888 + 0.0912 + 0.4712 = 1 \][/tex]

Answer:

x = 0.16823611831

Step-by-step explanation:

ln (natural log), when multiplied by e, cancels it out.

Therefore ln*e2x = ln*1.4

2x = 0.33647223662

x = 0.16823611831

Answer: [tex]x=0.168[/tex]

Step-by-step explanation:

To solve the equation [tex]e^{2x}=1.4[/tex] you need to apply natural logarithm to both sides of the equation:

[tex]ln(e)^{2x}=ln(1.4)[/tex]

According to the logarithms property:

[tex]ln(b)^a=aln(b)[/tex]

Then, applying the property, you get:

[tex](2x)ln(e)=ln(1.4)[/tex]

You need to remember the following:

[tex]ln(e)=1[/tex]

Therefore:

[tex]2x(1)=ln(1.4)\\\\2x=ln(1.4)[/tex]

And finally, you must divide both sides of the equation by 2:

[tex]\frac{2x}{2}=\frac{ln(1.4)}{2}\\\\x=0.1682[/tex]

Rounded to the nearest thousand:

[tex]x=0.168[/tex]

Answer:

a^3b^4

Step-by-step explanation:

You have 3 a's, so that would be

[tex]a^{3}[/tex]

and you have 4 b's so that would be

[tex]b^4[/tex]

so putting it together gives you

[tex]a^3b^4[/tex]

[tex]|\Omega|=52\\|A|=4\\\\P(A)=\dfrac{4}{52}=\dfrac{1}{13}[/tex]

The probability of getting the card of 7 in a deck will be  1 / 13.

Probability is defined as the ratio of the number of favourable outcomes to the total number of outcomes in other words the probability is the number that shows the happening of the event.

Probability = Number of favourable outcomes / Number of sample

Given that a card is selected at random from a standard deck of playing cards. The probability of getting a number 7 will be,

There are four 7 cards in the deck of 52 cards.

Number of favourable outcomes = 4

Number of sample = 52

The probability is,

P = 4 / 52

P = 1 / 13

Therefore, the probability of getting the card of 7 in a deck will be  1 / 13.

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Answer:

The right statement is sin(J) = cos(L) ⇒ answer D

Step-by-step explanation:

* Lets describe the figure

- LKJ is a right triangle, where K is a right angle

∵ m∠K = 90°

∵ LJ is opposite to angle K

∴ LJ is the hypotenuse

∵ LJ = 219

∵ KJ = 178

- By using Pythagoras Theorem

∵ (LJ)² = (LK)² + (KJ)²

∴ (219)² = (LK)² + (178)² ⇒ subtract (178)² from both sides

∴ (LK)² = (219)² - (178)²

∴ (LK)² = 16277

∴ LK = √16277 = 127.58

* Lets revise how to find the trigonometry function

# sin Ф = opposite/hypotenuse

# cos Ф = adjacent/hypotenuse

# tan Ф = opposite/adjacent

∵ LK is the opposite side to angle J

∵ LJ is the hypotenuse

∵ sin(J) = LK/LJ

∵ LK = 127.58 , LJ = 219

∴ sin(J) = 127.58/219 = 0.583

∵ LK is the adjacent side to angle L

∵ LJ is the hypotenuse

∵ cos(L) = LK/LJ

∵ LK = 127.58 , LJ = 219

∴ cos(L) = 127.58/219 = 0.583

∴ sin(J) = cos(L)

* The right statement is sin(J) = cos(L)

Answer:

a) {{}, {a}}.

b) {{}, {a}, {b}, {a, b}}.

c) {{}, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, {4}, {1, 4}, {2, 4}, {1, 2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}.

Step-by-step explanation:

The power set of a set is the set of all subset of the set in question. The number of power sets (including the empty set) of a set with [tex]n[/tex] (where [tex]n \in \mathbb{Z}[/tex]) unique elements is [tex]2^{n}[/tex].

In other words, there shall be

This explanation shows how to find the power set using binary numbers (only 0 and 1.) (Credit: Mathsisfun.)

List all the binary numbers that are equivalent to decimals ranging from 0 to [tex]2 - 1 = 1[/tex].

[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0 \\ 1 & 1\end{array}[/tex].

Reverse the original set. Each digit in the binary number corresponds to a member of the original set (i.e. a letter in a) and b) or a number in c).) 0 means that the element is absent in the subset and 1 means that the element is present.

[tex]\begin{array}{c|l}a & \text{Element of the Power Set}\\ 0 & \{\}\\ 1 & \{a\}\end{array}[/tex].

The power set of a) thus contains:

Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]4 - 1 = 3[/tex].

[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 00 \\ 1 & 01 \\ 2 & 10 \\ 3 & 11\end{array}[/tex].

[tex]\begin{array}{cc|l}b & a & \text{Element of the Power Set}\\ 0 & 0 & \{\}\\ 0 & 1 & \{a\}\\ 1 & 0 & \{b\} \\ 1 & 1 & \{a, b\}\end{array}[/tex].

The power set of b) thus contains:

Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]16 - 1 = 15[/tex].

[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0000 \\ 1 & 0001 \\ 2 & 0010 \\ 3 & 0011\\4 & 0100 \\ 5 & 0101\\ 6 & 0110 \\ 7 & 0111\\ 8 & 1000\\ 9 & 1001\\ 10 & 1010\\ 11 & 1011\\ 12& 1100 \\13 & 1101 \\ 14 & 1110\\ 15 & 1111 \end{array}[/tex].

[tex]\begin{array}{cccc|l}4 & 3 & 2 &1& \text{Element of the Power Set}\\ 0 & 0 & 0 & 0 &\{\}\\ 0 & 0 & 0 & 1 & \{1\}\\ 0 & 0 & 1 & 0 & \{2\} \\ 0 & 0 &1 & 1 & \{1, 2\} \\ 0 & 1 & 0 & 0 & \{3\} \\ 0 & 1 & 0 & 1& \{1, 3\}\\ 0 & 1 & 1 & 0& \{2, 3\}\\ 0 & 1 & 1 & 1 & \{1, 2, 3\} \\ 1 & 0 & 0 & 0 & \{4\} \\ 1 & 0 & 0 & 1 & \{1, 4\}\\ 1& 0 & 1 &0&\{2, 4\}\\ 1 & 0 & 1 & 1 &\{1, 2, 4\}\\ 1 & 1 & 0 & 0 & \{3, 4\} \\ 1 & 1 & 0 & 1 & \{1, 3, 4\} \\ 1 & 1 & 1 & 0 & \{2, 3, 4\} \\ 1 & 1 & 1 & 1 & \{1, 2, 3, 4\}\end{array}[/tex].

The power set of c) thus contains:

Answer:IF each vendor has his own price or (ppower) so far every single vendor will have his own price.

Step-by-step explanation:

The graph show\ing the demand (D) and supply (S = MC) curves in the market for hot dogs indicate: Competitive market.

In a market were their is competition, when demand and supply curves intersect this indicate market equilibrium.

Based on the graph the market equilibrium price will be $1.50 per hot dog while on the other hand the market equilibrium quantity will be 250 hot dogs which  is the point were demand and supply intersect.

Inconclusion the market for hot dogs indicate: Competitive market.

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Answer: 0.008

Step-by-step explanation:

Given: Mean : [tex]\mu=1200[/tex]

Standard deviation : [tex]\sigma = 60[/tex]

Sample size : [tex]n=36[/tex]

The formula to calculate z-score is given by :_

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x= 1224, we have

[tex]z=\dfrac{1224-1200}{\dfrac{60}{\sqrt{36}}}=2.4[/tex]

The P-value = [tex]P(z>2.4)=1-P(z<2.4)=1-0.9918024=0.0081976\approx0.008[/tex]

Hence, the probability that the sample mean will be larger than 1224 =0.008

To fill the hemispherical tank with water through a hole in the base, the work required is 418.88 foot-pounds.

To calculate the work required to fill the hemispherical tank with water through a hole in the base, we can use the concept of work done against gravity.

The volume of the tank can be calculated using the formula for the volume of a hemisphere, which is (2/3)πr^3. In this case, the radius is given as 2 feet.

The weight of the water can be found by multiplying the volume by the weight-density of water, which is 62.4 pounds per cubic foot.

The work done is then the weight of the water multiplied by the height it is lifted, which is equal to the radius of the hemisphere.

So, the work required to fill the tank with water is (2/3)π(2^3)(62.4)(2) = 418.88 foot-pounds.

The work required to fill the hemispherical tank with water through a hole in the base is [tex]$\frac{1248\pi}{5}$[/tex] foot-pounds.

To solve this problem, we need to calculate the work done against gravity to fill the tank with water. The work done to lift a small layer of water at a height [tex]$h$[/tex] is given by the force required to lift it times the height  [tex]$h$[/tex]. The force is the weight of the water, which is the volume of the water times its weight-density.

Let's break down the tank into thin horizontal slices. Each slice has a thickness [tex]$dh$[/tex] and is at a height [tex]$h$[/tex] from the base. The radius of each slice is given by [tex]$r = \sqrt{2^2 - h^2}$[/tex], where [tex]$2$[/tex] feet is the radius of the tank.

The volume of each slice is the area of the slice times its thickness [tex]$dh$[/tex]. The area of the slice is [tex]$\pi r^2$[/tex], so the volume [tex]$dV$[/tex] is [tex]\pi (2^2 - h^2) dh[/tex].

The weight of the water in each slice is the volume times the weight-density of water, which is [tex]$62.4$[/tex] pounds per cubic foot. Therefore, the weight of the water in each slice is [tex]$62.4\pi (2^2 - h^2) dh$[/tex].

The work done to lift this slice to the height [tex]$h$[/tex] is the weight of the water times the height [tex]$h$[/tex], which is [tex]$62.4\pi h (2^2 - h^2) dh$[/tex].

To find the total work done to fill the tank, we integrate this expression from [tex]$h = 0$[/tex]  to[tex]$h = 2$[/tex] feet:

[tex]\[ W = \int_{0}^{2} 62.4\pi h (2^2 - h^2) dh \] \[ W = 62.4\pi \int_{0}^{2} h (4 - h^2) dh \] \[ W = 62.4\pi \int_{0}^{2} (4h - h^3) dh \] \[ W = 62.4\pi \left[ 2h^2 - \frac{h^4}{4} \right]_{0}^{2} \] \[ W = 62.4\pi \left[ 2(2)^2 - \frac{(2)^4}{4} \right] - 62.4\pi \left[ 2(0)^2 - \frac{(0)^4}{4} \right] \] \[ W = 62.4\pi \left[ 8 - 4 \right] \] \[ W = 62.4\pi \tims 4 \] \[ W = 249.6\pi \] \[ W = \frac{1248\pi}{5} \][/tex]

Therefore, the work required to fill the hemispherical tank with water through a hole in the base is [tex]$\frac{1248\pi}{5}$[/tex] foot-pounds.

Answer:

In 20 years, the population will be about 80.3 thousand people

Step-by-step explanation:

If our first time is 0 and the population that goes along with that time is 70,000, we have a coordinate point where x is the time (0), and y is the population at that time (70).  Our next time is 10 years later, when the population is 75,000.  The coordinate point for that set of data is (10, 75).  Now we will use those 2 points in the standard form of an exponential equation to write the model for this particular situation.  

Exponential equations are of the form

[tex]y=a(b)^x[/tex]

where x and y are the coordinates from our points, one at a time; a is the initial value, and b is the growth rate.  Filling in an equation with the first set of data:

[tex]70=a(b)^0[/tex]

Anything raised to the power of 0 = 1, so b to the power of 0 = 1 and we simply have that a = 70.

Now we use that value of a along with the x and y from the next coordinate pair to solve for b:

[tex]75=70(b)^{10}[/tex]

Begin by dividing both sides by 70 to get

[tex]1.071428571=b^{10}[/tex]

Undo the power of 10 on the right by taking the 10th root of both sides:

[tex](1.071428571)^{\frac{1}{10}}=(b^{10})^{\frac{1}{10}}[/tex]

On the right side we simply have b now, and on the left we have

1.006923142=b

Now we have a and b to write the model for this situation:

[tex]y=70(1.006923142)^x[/tex]

We need to find y, the population, in x = 20 years:

[tex]y=70(1.006923142)^{20}[/tex]

Raise the parenthesis to the 20th power giving you

y = 70(1.147959784) and

y = 80.3 thousand people

Final answer:

The exponential growth function that fits the given data is y(t) = a * (1 + r)^t. Using this function, we can find the population 20 years from now.

Explanation:

The exponential growth function that fits the given data is:

y(t) = a * (1 + r)^t

where:

To find the growth rate per year, we can use the formula: r = (P/P0)^(1/t) - 1

Given that the population is projected to be 75,000 in 10 years, we can substitute these values into the formula to find the growth rate:

r = (75,000/70,000)¹/¹⁰ - 1 ≈ 0.035

The exponential growth function becomes:

y(t) = 70,000 * (1 + 0.035)^t

To find the population 20 years from now, we can substitute t = 20 into the exponential growth function:

y(20) = 70,000 * (1 + 0.035)²⁰ ≈ 95,212

Answer:

20m/s

Step-by-step explanation:

This can be solved using the acceleration / velocity equations.

Specifically,

v² = u² + 2as

Where

v = final velocity = what we need to find

u = initial velocity = given as 10.0m/s

a = acceleration = given as 0.5m/s²

s = distance = 300m

Hence,

v² = 10² + (2)(0.5) (300)

    = 100 + 300

    =400

v = √400 = 20 m/s

Jessica has a total of 22,275 different possible schedules to choose from for her next semester given the number of sections for each class and assuming there are no time conflicts.

Jessica is creating her semester schedule and there are 15 sections of English 101, 9 sections of Spanish 102, 11 sections of Biology 102, and 15 sections of College Algebra. To figure out how many different possible schedules are available, we need to multiply the number of sections for each class.

Therefore, the total number of different possible schedules Jessica can choose is calculated as follows:

15 (English 101) * 9 (Spanish 102) * 11 (Biology 102) * 15 (College Algebra) = 22,275 possible schedules.

This is under the assumption that there are no time conflicts between the different classes.

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