this is formula manipulation, I'd appreciate if the steps were provided. only respond if you know how to get the answer, thank you​

Answer:

112.77 gallons of water

Step-by-step explanation:

We will calculate the area by Trapezoid formula :

[tex]A=\frac{a+b}{2}\times h\times l[/tex]

Given Base a = 12 inches  = 1 feet

          Base b = 3 inches  =  0.25 feet

          height  = 8 inches  = 0.67 feet

          length  = 36 feet    = 36 feet

[tex]Area=\frac{1+2.5}{2}\times 0.67\times 36[/tex]

= 0.625 × 0.67 × 36

= 15.075 cubic feet.

As we know 1 cubic feet = 7.48052 per liquid gallon

Therefore, 15.075 cubic feet = 15.075 × 7.48052

                                               = 112.768839 ≈ 112.77 liquid gallon

When full it will hold 112.77 gallons of water

Answer:

Simple interest=4500

maturity value= 10500

Step-by-step explanation:

Given: Principal P = 6000

Rate % R = 5%

time T = 15 years

we know that simple interest [tex]SI=\frac{PRT}{100}[/tex]

⇒[tex]SI=\frac{6000\times5\times15}{100}[/tex]

on calculating we get SI = 4500

and maturity value = Principal amount + Simple interest

maturity value = 6000+4500 = 10500

hence the final answers are

Simple interest= 4500

and maturity value= 10500

hope this helps!!

Answer:

The center is the point (3,1) and the radius is 3 units

Step-by-step explanation:

we know that

The equation of a circle in standard form is equal to

[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]

we have

[tex]x^{2}+y^{2}-6x-2y+1=0[/tex]

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

[tex](x^{2}-6x)+(y^{2}-2y)=-1[/tex]

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

[tex](x^{2}-6x+9)+(y^{2}-2y+1)=-1+9+1[/tex]

[tex](x^{2}-6x+9)+(y^{2}-2y+1)=9[/tex]

Rewrite as perfect squares

[tex](x-3)^{2}+(y-1)^{2}=9[/tex]

The center is the point (3,1) and the radius is 3 units

The mean/expected value is

[tex]E[X]=\displaystyle\sum_x x\,P(X=x)=0P(X=0)+1P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)[/tex]

[tex]\implies E[X]=\boxed{1.987}[/tex]

d. The standard deviation is the square root of the variance, which itself is

[tex]V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]

We have

[tex]E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=0^2P(X=0)+1^2P(X=1)+2^2P(X=2)+3^2P(X=3)+4^2P(X=4)[/tex]

[tex]\implies E[X^2]=5.577[/tex]

Then the variance is

[tex]V[X]=5.577-1.987^2\approx1.629[/tex]

and so the standard deviation is

[tex]\sqrt{V[X]}\approx\boxed{1.276}[/tex]

e. We know this immediately from the table:

[tex]P(X=3)=\boxed{0.289}[/tex]

f. A parent can be involved in either 3 or 4 activities, but not simultaneously 3 and 4 activities (i.e. these events are disjoint), so

[tex]P(X=3\text{ or }X=4)=P(X=3)+P(X=4)=\boxed{0.39}[/tex]

Applying statistical concepts, we have that:

c) The mean is of 1.987 activities.

d) The standard deviation is of 1.67 activities.

e) The probability is 0.289.

f) The probability is 0.39.

The probability distribution is given by:

[tex]P(X = 0) = 0.216[/tex]

[tex]P(X = 1) = 0.072[/tex]

[tex]P(X = 2) = 0.322[/tex]

[tex]P(X = 3) = 0.289[/tex]

[tex]P(X = 4) = 0.101[/tex]

Item c:

The mean is given by the sum of the multiplications of each outcome by it's probability, thus:

[tex]E(X) = 0.216(0) + 0.072(1) + 0.322(2) + 0.289(3) + 0.101(4) = 1.987[/tex]

The mean is of 1.987 activities.

Item d:

The standard deviation is given by the square root of the sum of the squares of each outcome subtracted by the mean, multiplied by it's probability, thus:

[tex]\sqrt{V(X)} = \sqrt{0.216(0-1.987)^2 + 0.072(1-1.987)^2) + 0.322(2-1.987)^2) + ...}[/tex]

[tex]\sqrt{V(X)} = 1.67[/tex]

The standard deviation is of 1.67 activities.

Item e:

This probability is P(X = 3), thus 0.289.

Item f:

This probability is:

[tex]p = P(X = 3) + P(X = 4) = 0.289 + 0.101 = 0.39[/tex]

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With 22 packed cheese sandwiches made, bread is the 'limiting' ingredient, and we have three slices of cheese in excess.

The number of sandwiches that can be made depends on the availability of the critical ingredients needed to complete a sandwich: 2 slices of bread and three slices of cheese. To calculate the number of sandwiches, we can determine how many sandwiches each ingredient batch could provide, then find the minimum of those two, as the 'limiting' factor will be the ingredient that finishes first.

For bread, we have 44 pieces/ 2 pieces per sandwich = 22 sandwiches. With cheese, we have 69 pieces/ 3 pieces per sandwich = 23 sandwiches. Therefore, we can fully make a maximum of 22 sandwiches because we would run out of bread first. This means the bread is the 'limiting' ingredient, while cheese is in excess. With 22 sandwiches made, we would use three slices of cheese per sandwich for 66 slices used, leaving three slices of cheese remaining. The quantity in excess is three slices of cheese.

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There are 0 slices of bread in excess, and there are 3 slices of cheese in excess.

Let's start by calculating the maximum number of sandwiches that can be made using 44 slices of bread:

Each sandwich requires 2 slices of bread.

So, the maximum number of sandwiches that can be made using 44 slices of bread is 44/2 = 22 sandwiches.

Now, let's calculate the maximum number of sandwiches that can be made using 69 slices of cheese:

Each sandwich requires 3 slices of cheese.

So, the maximum number of sandwiches that can be made using 69 slices of cheese is 69/3 = 23 sandwiches.

Now, we compare the results:

With 44 slices of bread, we can make 22 sandwiches.

With 69 slices of cheese, we can make 23 sandwiches.

Since we can only make 22 sandwiches due to the limitation of the bread, the bread is the limiting ingredient.

The quantity of the ingredient in excess can be found by subtracting the number of sandwiches that can be made using the limiting ingredient from the total number of slices of that ingredient.

For the bread:

Quantity of bread in excess = Total slices of bread - (Number of sandwiches × Slices of bread per sandwich)

Quantity of bread in excess = 44 - (22 *2)

= 44 - 44

= 0

For the cheese:

Quantity of cheese in excess = Total slices of cheese - (Number of sandwiches × Slices of cheese per sandwich)

Quantity of cheese in excess = 69 - (22 *s 3)

= 69 - 66

= 3

So, there are 0 slices of bread in excess, and there are 3 slices of cheese in excess.

Answer:

$ 1.96

Step-by-step explanation:

Number of spots with outcome of $1 = 9

Number of spots with outcome of $2 = 18

Number of spots with outcome of $10 = 1

Total number of spots = 28

Probability that ball will land on $1 = [tex]\frac{9}{28}[/tex]

Probability that ball will land on $2 = [tex]\frac{18}{28}[/tex]

Probability that ball will land on $10 = [tex]\frac{1}{28}[/tex]

The amount that player should expect to win on average in equal to expected value of the game. Expected value is calculated as the summation of product of probabilities with their respective outcomes.

i.e. for this case:

Expected Value will be:

[tex](1 \times \frac{9}{28})+(2 \times \frac{18}{28})+(10 \times \frac{1}{28})\\\\ =1.96[/tex]

This means, on average the player should expect to win $ 1.96

Probability is the ratio of the favorable event to the total number of events. The average amount is $1.96.

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen.

A new game is being introduced at the Hard Rock Cafe.

A ball is spun around a wheel until it comes to rest in one of many spots. Whatever is listed in that spot will be the player's winnings.

If the wheel following spots

Number of spots labeled $1 = 9

Number of spots labeled $2 = 18

Number of spots labeled $10 = 1

The total number of spots will be

Total spots = 28

The probability of the ball landing on $1 will be

[tex]\rm P(\$1) = \dfrac{9}{28}[/tex]

The probability of the ball landing on $2 will be

[tex]\rm P(\$2) = \dfrac{18}{28}[/tex]

The probability of the ball landing on $10 will be

[tex]\rm P(\$10) = \dfrac{1}{28}[/tex]

The amount that player should expect to win on average is equal to the expected value of the game will be

[tex]\rm Average\ amount = 1*\dfrac{9}{28} + 2*\dfrac{18}{28} + 10*\dfrac{1}{28}\\\\\\Average \ amount = 1.96[/tex]

More about the probability link is given below.

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Answer:

Step-by-step explanation:

What is a regular tessellation?

A regular tessellation is a pattern made by repeating a regular polygon. In simpler words regular tessellations are made up entirely of congruent regular polygons all meeting vertex to vertex.

How many regular tessellation are possible?

There are only 3 regular tessellation.

1. Triangle

2. Square

3. Hexagon

Why aren't there infinitely many regular tessellations?

Not more than 3 regular tessellations are possible because the sums of the interior angles are either greater than or less than 360 degrees....

Final answer:

A regular tessellation is a repeating pattern of a regular polygon that fills a plane without gaps or overlaps, with only three possible: equilateral triangles, squares, and regular hexagons. There aren't infinitely many because a tessellation requires the angles at a vertex to sum to 360 degrees, a condition only satisfied by these three shapes.

Explanation:

A regular tessellation is a pattern made by repeating a regular polygon to fill a plane without any gaps or overlaps. There are exactly three regular tessellations possible, which are constituted by:

There aren't infinitely many regular tessellations because for a regular polygon to tessellate, the sum of the angles at a vertex where polygons meet must be exactly 360 degrees.

This criterion is satisfied only by triangles (each angle is 60 degrees), squares (each angle is 90 degrees), and hexagons (each angle is 120 degrees).

Polygons with more sides have larger interior angles, such that the sum exceeds 360 degrees, preventing them from tessellating regularly.

Answer:

The equation to find the number of fish after t years where y is the number of fish is:

y(t) = 160×3^(t) ( t <= 2.93 assuming that the maximum number of fish is 4000 )

Therefore when y(t) = 2000 2000 = 160×3^(t)

3^(t) = 25/2 log 3 (3^(t)) = log 3 (25/2) t = 2.29 years.

Using the logistic equation, we have that:

a)

The equation is:

[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

b)

It will take 2.68 years for the population to increase to 2000.

The logistic equation is:

[tex]P(t) = \frac{K}{1 + Ae^{-kt}}[/tex]

With:

[tex]A = \frac{K - P(0)}{P(0)}[/tex]

The parameters are:

In this problem:

Then:

[tex]A = \frac{4000 - 160}{160} = 24[/tex]

Thus:

[tex]P(t) = \frac{4000}{1 + 24e^{-kt}}[/tex]

Item a:

Tripled during the first year, thus [tex]P(1) = 3P(0) = 3(160) = 480[/tex].

This is used to find k.

[tex]480 = \frac{4000}{1 + 24e^{-k}}[/tex]

[tex]480 + 11520e^{-k} = 4000[/tex]

[tex]e^{-k} = \frac{3520}{11520}[/tex]

[tex]\ln{e^{-k}} = \ln{\frac{3520}{11520}}[/tex]

[tex]k = -\ln{\frac{3520}{11520}}[/tex]

[tex]k = 1.1856[/tex]

Thus, the equation is:

[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

Item b:

This is t for which P(t) = 2000, thus:

[tex]P(t) = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

[tex]2000 = \frac{4000}{1 + 24e^{-1.1856t}}[/tex]

[tex]\frac{1}{1 + 24e^{-1.1856t}} = 0.5[/tex]

[tex]0.5 + 12e^{-1.1856t} = 1[/tex]

[tex]e^{-1.1856t} = \frac{1}{24}[/tex]

[tex]\ln{e^{-1.1856t}} = \ln{\frac{1}{24}}[/tex]

[tex]t = -\frac{\ln{\frac{1}{24}}}{1.1856t}[/tex]

[tex]t = 2.68[/tex]

It will take 2.68 years for the population to increase to 2000.

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Answer:

  y = 2(x + 2)^2 – 5

Step-by-step explanation:

When y = ax^2 +bx +c is written in vertex form, it becomes ...

  y = a(x +b/(2a))^2 +(c -b^2/(4a))

The constant term in the squared binomial is b/(2a) = 8/(2(2)) = +2. Only one answer choice matches:

  y = 2(x +2)^2 -5

Answer:

a < -12

Step-by-step explanation:

Isolate the variable, a. Treat the < as a equal sign, what you do to one side, you do to the other. Subtract 2 from both sides:

a + 2 < -10

a + 2 (-2) < -10 (-2)

Simplify:

a < -10 - 2

a < -12

a < -12 is your answer.

~

Answer:

[tex]\huge \boxed{a<-12}\checkmark[/tex]

Step-by-step explanation:

Subtract by 2 both sides of equation.

[tex]\displaystyle a+2-2<-10-2[/tex]

Simplify, to find the answer.

[tex]-10-2=-12[/tex]

[tex]\displaystyle a<-12[/tex], which is our answer.

Answer: [tex] 495[/tex]

Step-by-step explanation:

We know that if the population is normally distributed with mean [tex]\mu[/tex]  and standard deviation [tex]\sigma[/tex], then the sampling distribution of the sample mean , [tex]\overline{x}[/tex] is also normally distribution with :-

Mean : [tex]\mu_{\overlien{x}}=\mu[/tex]

Given : The scores on the Critical Reading part of the SAT exam in a recent year were roughly Normal with mean [tex]\mu= 495[/tex]

Then , the mean of the average scores you get will be close to [tex] 495[/tex]

We can express the recurrence,

[tex]\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}[/tex]

in matrix form as

[tex]\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}[/tex]

By substitution,

[tex]\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}[/tex]

and continuing in this way we would find that

[tex]\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]

Diagonalizing the coefficient matrix gives us

[tex]\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}[/tex]

which makes taking the [tex](n-2)[/tex]-th power trivial:

[tex]\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}[/tex]

[tex]\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}[/tex]

So we have

[tex]\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]

and in particular,

[tex]a_n=\dfrac{29\left(2(-5)^{n-1}+5\cdot2^{n-1}\right)-47\left(-(-5)^{n-1}+2^{n-1}\right)}7[/tex]

[tex]a_n=\dfrac{105(-5)^{n-1}+98\cdot2^{n-1}}7[/tex]

[tex]a_n=15(-5)^{n-1}+14\cdot2^{n-1}[/tex]

[tex]\boxed{a_n=-3(-5)^n+7\cdot2^n}[/tex]

To find the probability that more than 1 student will have their automobile stolen during the current semester, we can use a Poisson distribution. The average number of automobile thefts per semester is 7. By calculating 1 minus the probability of 0 or 1 thefts, we find that the probability of more than 1 theft is approximately 0.9991.

To find the probability that more than 1 student will have their automobile stolen during the current semester, we can use a Poisson distribution. The Poisson distribution models the number of events that occur in a fixed interval of time or space. In this case, the average number of automobile thefts per semester is given as 7. We want to find the probability that more than 1 student will have their automobile stolen, so we need to calculate 1 minus the probability that 0 or 1 student will have their automobile stolen.

The probability mass function of a Poisson distribution is given by P(X=k) = e^{-λ} * (λ^k) / k! where λ is the average number of events. In this case, λ = 7.

So, the probability that more than 1 student will have their automobile stolen is P(X>1) = 1 - (P(X=0) + P(X=1)).

To calculate P(X=0) and P(X=1), we can substitute k=0 and k=1 into the probability mass function and sum them up.

P(X=0) = e^{-7} * (7^0) / 0! = e^{-7}

P(X=1) = e^{-7} * (7^1) / 1! = 7e^{-7}

Therefore, P(X>1) = 1 - (e^{-7} + 7e^{-7}).

Rounding this answer to four decimal places, we get P(X>1) ≈ 0.9991.

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The number of different possible 7-number selections from a pool of 48, where order does not matter, can be calculated using the mathematical concept of combinations. The formula to calculate the total number of combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of options and k is the number of options selected.

In the LOTTO game you described, you must select 7 numbers from a pool of 48, and the order of the numbers does not matter. This is a problem of combinations in mathematics. The formula for combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of options, k is the number of options selected, and '!' denotes a factorial. In this case, n=48 (the numbers 1 through 48) and k=7 (the seven numbers you select).

By plugging these values into the formula, we can calculate the total number of different selections possible: C(48, 7) = 48! / [7!(48-7)!]. This calculation would give us the total number of combinations of 7 numbers that can be selected from a pool of 48, which represents all the different possible LOTTO selections. It should be remembered that factorials such as 48! or 7! represent a product of an integer and all the integers below it, down to 1.

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There are 85,900,584 different selections possible in the LOTTO game.

To calculate the number of different selections possible in the LOTTO game, we need to focus on the concept of permutations.

With 48 numbers to choose from and 7 numbers to be selected, the number of permutations can be calculated using the formula P(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected.

In this case, n = 48 and r = 7.

Using the formula, we have P(48, 7) = 48! / (48-7)! = (48 * 47 * 46 * 45 * 44 * 43 * 42) / (7 * 6 * 5 * 4 * 3 * 2 * 1)

= 85,900,584 different selections possible.

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Answer:

The value of r(s(3)) = -21

Step-by-step explanation:

It is given that,

r(x) = -2x + 1

s(x) = -x^2 + 2

To find the value of r(s(3))

s(x) = -x^2 + 2

s(3) = (-3)^2 + 2    [Substitute 3 instead of x]

 = 9 + 2

 = 11

Therefore s(3) = 11

r(x) = -2x + 1

r(s(3)) = r(11)   [Substitute 11 instead of x]

 =   -2(11) + 1

 = -22 + 1

  = -21

Therefore the value of r(s(3)) = -21

The answer is:

[tex]r(s(3))=15[/tex]

To solve the problem, first, we need to compose the functions, and then evaluate the obtained function. Composing function means evaluating a function into another function.

We have that:

[tex]f(g(x))=f(x)\circ g(x)[/tex]

From the statement we know the functions:

[tex]r(x)=-2x+1\\s(x)=-x^{2}+2[/tex]

We need to evaluate the function "s" into the function "r", so:

[tex]r(s(x))=-2(-x^2+2)+1\\\\r(s(x))=2x^{2}-4+1=2x^{2}-3[/tex]

Now, evaluating the function, we have:

[tex]r(s(3))=2(3)^{2}-3=2*9-2=18-3=15[/tex]

Have a nice day!

Answer:

33

Step-by-step explanation:

The mode is the number that sets up the most in an answer.

Remember, the stem is the number that is in the "tens" place value, while the leaves is the number in the "ones" place value.

Expand the stem-leaves:

10, 13, 16, 20, 21, 23, 27, 27, 28, 29, 30, 32, 33, 33, 33, 33, 38, 39, 46, 46, 46, 48

The number that shows up the most is 33, and is your answer.

~

Answer:

  6 cm by 17 cm

Step-by-step explanation:

The area is the product of the dimensions; the perimeter is double the sum of the dimensions.

So, we want to find two numbers whose product is 102 and whose sum is 23.

  102 = 1·102 = 2·51 = 3·34 = 6·17

The last of these factor pairs has a sum of 23.

The dimensions are 6 cm by 17 cm.

Answer:

Step-by-step explanation:Find the slope of the line that passes through the points shown in the table.

The slope of the line that passes through the points in the table is

.

Using the hypergeometric distribution, the required probability is 0.0129.

Given that, the town had 7 employees over 50 years of age and 15 under 50.

Let, n₁=7 and n₂=15

Total employees in the town = 7+15  = 22

We have to find the probability that exactly 1 dismissed employee was over 50.

That means the remaining 8 dismissed employees are under 50.

Let the number of dismissed employees under 50 is denoted by x and above 50 denoted by y.

Then, using the hypergeometric distribution, the required probability can be calculated as:

[tex]P(x=1, y=8) = \dfrac{\binom{7}{1}+\binom{15}{8}}{\binom{22}{9}}[/tex]

[tex]P(x=1, y=8) = \dfrac{7+6435}{ 497420}[/tex]

[tex]P(x=1, y=8) = \dfrac{6442}{ 497420}[/tex]

[tex]P(x=1, y=8) = 0.0129[/tex]

Hence, the probability that exactly 1 employee was over 50 is 0.0129.

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Answer:

The correct options are a,b,e and f.

Step-by-step explanation:

It is given that the nth terms of the series is defined as

[tex]T_n=3n+17[/tex]

Subtract 17 from both the sides.

[tex]T_n-17=3n[/tex]

Divide both sides by 3.

[tex]\frac{T_n-17}{3}=n[/tex]

The term Tₙ is a term of given series if n is a positive integer.

(a) The given term is 80.

[tex]n=\frac{80-17}{3}=21[/tex]

Since n is a positive integer, therefore 80 is a term of given series.

(b) The given term is 170.

[tex]n=\frac{170-17}{3}=51[/tex]

Since n is a positive integer, therefore 170 is a term of given series.

(c) The given term is 217.

[tex]n=\frac{217-17}{3}=66.67[/tex]

Since n is not a positive integer, therefore 217 is a term of given series.

(d) The given term is 312.

[tex]n=\frac{312-17}{3}=98.33[/tex]

Since n is not a positive integer, therefore 312 is a term of given series.

(e) The given term is 278.

[tex]n=\frac{278-17}{3}=87[/tex]

Since n is a positive integer, therefore 278 is a term of given series.

(f) The given term is 3566.

[tex]n=\frac{3566-17}{3}=1183[/tex]

Since n is a positive integer, therefore 3566 is a term of given series.

Thus, the correct options are a, b, e and f.

Answer:

(a) $13811.68

(b) $13914.30

(c) $14004.55

(d) $14022.54

(e) $14023.15

Step-by-step explanation:

Since, the amount formula in compound interest,

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]

Where,

P = Principal amount,

r = annual rate,

n = number of periods,

t = number of years,

Here, P = $ 9,400, r = 8% = 0.08, t = 5 years,

If the amount is compounded annually,

n = 1,

Hence, the amount of investment would be,

[tex]A=9400(1+\frac{0.08}{1})^5=9400(1.08)^5=\$ 13811.6839219\approx \$ 13811.68[/tex]

(a) If the amount is compounded annually,

n = 1,

The amount of investment would be,

[tex]A=9400(1+\frac{0.08}{1})^5=9400(1.08)^5=\$ 13811.6839219\approx \$ 13811.68[/tex]

(b) If the amount is compounded semiannually,

n = 2,

The amount of investment would be,

[tex]A=9400(1+\frac{0.08}{2})^{10}=9400(1.04)^{10}=\$13914.2962782\approx \$ 13914.30[/tex]

(c) If the amount is compounded Monthly,

n = 12,

The amount of investment would be,

[tex]A=9400(1+\frac{0.08}{12})^{60}=9400(1+\frac{1}{150})^{60}=\$ 14004.549658\approx \$ 14004.55[/tex]

(d) If the amount is compounded Daily,

n = 365,

The amount of investment would be,

[tex]A=9400(1+\frac{0.08}{365})^{365\times 5}=9400(1+\frac{2}{9125})^{1825}=\$ 14022.5375476\approx \$ 14022.54[/tex]

(e) Now, the amount in compound continuously,

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

Where, P = principal amount,

r = annual rate,

t = number of years,

So, the investment would be,

[tex]A=9400 e^{0.08\times 5}=9400 e^{0.4}=\$14023.1521578\approx \$14023.15[/tex]

Step-by-step explanation:

There's a lot of information here, so first things first, let's get organized.

Let's start by assigning variable names to each group.  We want the variables to be short but easy to understand.

For example, let's say the number of female seniors on the dean's list is FSD (F for female, S for senior, and D for dean's list).

FSD = 31

Sticking to this naming scheme:

FD = 62

MSD = 45

FS = 87

MS = 96

F = 275

MD = 88

M = 227

Now we can begin.

a) We want to know how many seniors there are.  So all we have to do is add up all the variables with an S in them.

FSD + MSD + FS + MS

31 + 45 + 87 + 96

259

b) We want to know how many women there are.  So add up all the variables with an F in them.

FSD + FD + FS + F

31 + 62 + 87 + 275

455

c) We want to know how many are on the dean's list.  So add up all the variables with a D in them.

FSD + FD + MSD + MD

31 + 62 + 45 + 88

226

d) Now we want to know how many are seniors AND on the dean's list.  So add up all the variables that have both an S and a D.

FSD + MSD

31 + 45

76

e) We want to know how many female seniors there are, so add up all the variables with both an F and an S.

FSD + FS

31 + 87

118

f) We want to know how many women were on the dean's list, so add up all the variables with both an F and a D.

FSD + FD

31 + 62

93

g) Finally, we want to know how many students there are total.  So add up all the variables.

FSD + FD + MSD + FS + MS + F + MD + M

62 + 45 + 87 + 96 + 275 + 88 + 227

880

Using arithmetic operations based on the provided data, we deduced the total number of seniors, women, students on the dean's list, and overall students at the college, among other specific categorizations.

To solve these problems, we will first interpret the given data, then perform simple arithmetic operations based on the information provided to answer each part of the question.

Answer:

[tex]x+\sqrt{3}y=8[/tex]

Step-by-step explanation:

Given equation of curve,

[tex]y^2+x^2=16[/tex]

[tex]\implies y^2=16-x^2[/tex]

Differentiating with respect to x,

[tex]2y\frac{dy}{dx}=-2x[/tex]

[tex]\implies \frac{dy}{dx}=-\frac{x}{y}[/tex]

Since, the tangent line of the curve contains the point (2, 2√3),

Thus, the slope of the tangent line,

[tex]m=\left [ \frac{dy}{dx} \right ]_{(2, 2\sqrt{3})}=-\frac{1}{\sqrt{3}}[/tex]

Hence, the equation of tangent line would be,

[tex]y-2\sqrt{3}=-\frac{1}{\sqrt{3}}(x-2)[/tex]

[tex]\sqrt{3}y-6=-x+2[/tex]

[tex]\implies x+\sqrt{3}y=8[/tex]

Answer:

[tex]14878.04878miles/hours^2[/tex]

Step-by-step explanation:

Let's find a solution by understanding the following:

The acceleration rate is defined as the change of velocity within a time interval, which can be written as:

[tex]A=(Vf-Vi)/T[/tex] where:

A=acceleration rate

Vf=final velocity

Vi=initial velocity

T=time required for passing from Vi to Vf.

Using the problem's data we have:

Vf=65miles/hour

Vi=6miles/hour

T=14.8seconds

Using the acceleration rate equation we have:

[tex]A=(65miles/hour - 6miles/hour)/14.8seconds[/tex], but look that velocities use 'hours' unit while 'T' uses 'seconds'.

So we need to transform 14.8seconds into Xhours, as follows:

[tex]X=(14.8seconds)*(1hours/60minutes)*(1minute/60seconds)[/tex]

[tex]X=0.0041hours[/tex]

Using X=0.0041hours in the previous equation instead of 14.8seconds we  have:

[tex]A=(65miles/hour - 6miles/hour)/0.0041hours[/tex]

[tex]A=(61miles/hour)/0.0041hours[/tex]

[tex]A=(61miles)/(hour*0.0041hours)[/tex]

[tex]A=61miles/0.0041hours^2[/tex]

[tex]A=14878.04878miles/hours^2[/tex]

In conclusion, the acceleration rate is [tex]14878.04878miles/hours^2[/tex]

Answer:

68600 will there be at​ midnight ( approx )

Step-by-step explanation:

Let P shows the population of the bacteria,

Since, the number of bacteria in a culture grew at a rate proportional to its size,

[tex]\implies \frac{dP}{dt}\propto P[/tex]

[tex]\frac{dP}{dt}=kP[/tex]

Where, k is the constant of proportionality,

[tex]\frac{dP}{P}=kdt[/tex]

[tex]\int \frac{dP}{P}=\int kdt[/tex]

[tex]ln P=kt + C_1[/tex]

[tex]P=e^{kt+C_1}[/tex]

[tex]P=e^{kt}.e^{C_1}=C e^{kt}[/tex]

Now, let the population of bacteria is estimated from 10:00 AM,

So, at t = 0, P = 4,000 ( given )

[tex]4000 = Ce^{0}[/tex]

[tex]\implies C=4000[/tex]

Now, at noon there are 4,600 bacterias,

That is, at t = 2, P = 4600

[tex]4600=Ce^{2k}[/tex]

[tex]4600 = 4000 e^{2k}[/tex]

[tex]\implies e^{2k}=\frac{4600}{4000}=1.15[/tex]

[tex]2k=ln(1.5)\implies k=\frac{ln(1.5)}{2}=0.202732554054\approx 0.203[/tex]

Hence, the equation that represents the population of bacteria after t hours,

[tex]P=4000 e^{0.203t} [/tex]

Therefore, the population of the bacteria at midnight ( after 14 hours ),

[tex]P=4000 e^{0.203\times 14}=4000 e^{2.842}= 68600.1252903\approx 68600[/tex]

There are 9,072 different routes possible for the delivery driver.

To find the number of different routes possible, we can use the concept of permutations. Since the driver has to deliver to 7 out of the 9 remaining locations, we can calculate the number of ways to choose 7 out of 9 and then multiply it by the number of ways to arrange those 7 locations. The formula for permutations is P(n, r) = n! / (n - r)!. In this case, n = 9 and r = 7.

P(9, 7) = 9! / (9 - 7)! = 9! / 2! = 9 × 8 × 7 × 6 × 5 × 4 × 3 = 9,072

Therefore, there are 9,072 different routes possible for the delivery driver.

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To find the number of different routes possible, we can use the combination formula: C(n, k) = n! / (k!(n-k)!). Plugging in the values, we get C(9, 7) = 36. Therefore, there are 36 different routes possible.

To find the number of different routes possible, we can use the combination formula:

C(n, k) = n! / (k!(n-k)!)

Where n is the total number of locations remaining (9) and k is the number of locations the driver can make deliveries to (7).

Plugging in the values, we get:

C(9, 7) = 9! / (7!(9-7)!)

= 9! / (7!2!)

= (9 * 8 * 7!)/(7! * 2!)

= (9 * 8)/(2 * 1)

= 36

Therefore, there are 36 different routes possible.

Answer:

15%

Step-by-step explanation:

[tex]y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}[/tex]

Divide both sides by [tex]\dfrac13y^{-2}(x)[/tex]:

[tex]3y^2y'-y^3=xe^x\ln x[/tex]

Substitute [tex]v(x)=y(x)^3[/tex], so that [tex]v'(x)=3y(x)^2y'(x)[/tex].

[tex]v'-v=xe^x\ln x[/tex]

Multiply both sides by [tex]e^{-x}[/tex]:

[tex]e^{-x}v'-e^{-x}v=x\ln x[/tex]

The left side can be condensed into the derivative of a product.

[tex](e^{-x}v)'=x\ln x[/tex]

Integrate both sides to get

[tex]e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C[/tex]

Solve for [tex]v(x)[/tex]:

[tex]v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]

Solve for [tex]y(x)[/tex]:

[tex]y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]

[tex]\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}[/tex]

Answer:

Step-by-step explanation:

Each Calculus text returns $10/2 = $5 per unit of shelf space. For History and Marketing texts, the respective numbers are $4/1 = $4 per unit, and $8/4 = $2 per unit. Using 1200 units of shelf space for 600 Calculus texts returns ...

  $5/unit × 1200 units = $6000 . . . profit

Any other use of units of shelf space will reduce profit.

The College should purchase only 600 Calculus textbooks to maximize College profits.

Data and Calculations:

Budget limit on the number of textbooks per semester = 700

Shelf space for the purchased textbooks = 1,200 units

                                                       Calculus      History      Marketing

Shelf-space occupied

by each textbook                              2                   1                  4

Profit per textbook                        $10                 $4               $8

Profit per shelf-space                    $5 ($10/2)    $4 ($4/1)     $2 ($8/4)

The highest profit per shelf space is $5 generated by Calculus.

The highest profit that the College can make over the purchase of used Calculus textbooks = $6,000 ($5 x 1,200) or ($10 x 600).

Thus, for the College to maximize its profits, it should purchase 600 Calculus textbooks, which will not exceed the textbook limit for the semester, and at the same time maximally utilize the shelf space available.

Related link: https://brainly.com/question/14293956

Answer:

0.1971 ( approx )

Step-by-step explanation:

Let X represents the event of weighing more than 20 pounds,

Since, the binomial distribution formula is,

[tex]P(x)=^nC_r p^r q^{n-r}[/tex]

Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Given,

The probability of weighing more than 20 pounds, p = 25% = 0.25,

⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

[tex]P(X<3) = P(X=0)+P(X=1)+P(X=2)[/tex]

[tex]=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}[/tex]

[tex]=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}[/tex]

[tex]=0.1971110499[/tex]

[tex]\approx 0.1971[/tex]