Students returning to a secondary school have to pass through a check point at the rate of 50 per though. The average time for checking is 35 seconds. The arrival rate and service rate follow a Poisson distribution. There is a delay and the principal is willing to include another check point to reduce the average time of checking to 28 seconds if the idle time is less than 10% and the average queue length is more than 10 students, check whether the introduction of an extra check point is justifiable.

Answer:

Using the ratio and proportion method, we find that on an average 1.875 rooms can be painted per gallon.

Step-by-step explanation:

Concept: Use the method of ratio and proportion.

In this method, if a relationship is given between two values, then using that relationship we can find another relationship with respect to another variable.

Given that for 2/5 gallons of paint, we can paint 3/4 of rooms. So, for one gallon of paint, divide 2/5 by 3/4.

2/5 gallons of paint = 3/4 rooms

1 gallon of paint = [tex]\frac{\frac{3}{4} }{\frac{2}{5} } = \frac{15}{8} =1.875[/tex] rooms

So, on an average they can paint 1.875 rooms per gallon.

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Answer:

area of a parallelogram equals to 1/2 x base x perpendicular height and area of a triangle equals to 1/2 base times height so you ought to search for the base in order to get the area of the triangle needed that's how I got the answer bye.

Answer:

A. y = 600x, the jet travels 12,000 miles in 20 hours

Explanation:

[tex]\sf slope: \dfrac{y_2 - y_1}{x_2- x_1} \ \ where \ (x_1 , \ y_1), ( x_2 , \ y_2) \ are \ points[/tex]

Take two points: (2, 1200), (4, 2400)

[tex]\sf slope : \dfrac{2400-1200}{4-2 } = 600[/tex]

Equation:

y - 1200 = 600(x - 2)

y - 1200 = 600x - 1200

y = 600x

Miles the Jet travels at 20 hours:

Miles the Jet travels at 22 hours:

Answer:

y = 600x, the jet travels 12,000 miles in 20 hours

Step-by-step explanation:

Slope-intercept form of a linear equation:

[tex]y = mx + b[/tex]

where:

To find the slope, define two points on the line and use the slope formula to find the slope:

[tex]\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1200-0}{2-0}=600[/tex]

From inspection of the graph, the y-intercept (where the line crosses the y-axis) is at (0, 0).  Therefore, the equation of the line is:

y = 600x

y is defined as the distance (in miles).  Therefore, to find the number of hours it takes for the jet to travel 12,000 miles, substitute y = 12000 into the found equation and solve for x:

[tex]\sf \implies 600x=12000[/tex]

[tex]\sf \implies x=\dfrac{12000}{600}[/tex]

[tex]\sf \implies x=\dfrac{120}{6}[/tex]

[tex]\implies \sf x=20[/tex]

Therefore, it takes the jet 20 hours to travel 12,000 miles.

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Answer:  [tex]\textsf{3955cm}[/tex]

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Given: [tex]\textsf{Tree = 4m and 25cm, Pole = 70cm shorter}[/tex]

Find: [tex]\textsf{The height of the pole}[/tex]

Solution: The first step that we must take is to convert the tree height to centimeters and after doing so we would just subtract 70 cm from that to get the pole height.

Convert to cm

Combine

Subtract 70 from the tree height

Using the information from the problem the height of the pole would be 3955cm.

Answer:

a, b, d

Step-by-step explanation:

its one of those

Answer:

A, the first one

Answer:

  16x² +48x +36

Step-by-step explanation:

A perfect square trinomial is of the form ...

  (a +b)² = a² +2ab +b²

We want to match this form.

__

Comparing the known terms, we see ...

  16x² = a²   ⇒   a = 4x

  36 = b²   ⇒   b = 6

The missing term is the linear term:

  2ab = 2(4x)(6) = 48x

  ? = 48

Answer:

[tex]4\sqrt{26}[/tex]

Step-by-step explanation:

The distance formula is [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex].

Thus we can plug the numbers in:

[tex]\sqrt{(-2-(-6))^2+(10-(-10))^2}[/tex]

[tex]\sqrt{(4)^2+(20)^2}[/tex]

[tex]\sqrt{16+400}[/tex]

[tex]\sqrt{416}[/tex]

[tex]4\sqrt{26}[/tex]

Answer:

Step-by-step explanation:

→ Find the gradient

( -3 --4 ) ÷ ( 0 --3) = 1/3

→ Write into format

y = 1/3x + c

→ Substitute in a coordinate pair

-3 = 0 + c

→ Solve for c

c = -3

→ Write into format

y = 1/3x - 3

Answer:

20-2h

Step-by-step explanation:

20 is the original amount. The variable stands for the number of hours. You are TAKING AWAY 2 degrees every hour so you would subtract.

Answer:

I believe that it should be Y; Height of an emperor penguin on Antarctica

Answer:

Z = "results of flipping two coins"

Step-by-step explanation:

A discrete variable is one which can take only certain values and usually represents values that are counted. The result of flipping two coins can only take a certain number of values of heads or tails (1 head and 1 tail, two heads, or two tails); therefore it is a discrete variable.

A continuous variable is one which can take any value, and usually represents values that are measured. The rest of the options provided (W, X, and Y) are all measurements, and are therefore continuous, not discrete.

Also, among the options presented, option Z is the only one which is "random".

Answer:

  (a)  y = 5/12x and y = -5/12x

Step-by-step explanation:

Given the equation of a hyperbola 25x² – 144y² + 3,600 = 0, you want the equations of the asymptotes.

Dividing by 3600 and doing a little rearrangement, we have ...

  (y/5)² -(x/12)² = 1 . . . . . standard form equation of a hyperbola

The equation of the asymptotes is ...

  (y/5)² -(x/12)² = 0

Solving for y gives ...

  (y/5)² = (x/12)²

  y/5 = ±x/12 . . . . . . take the square root

  y = ±5/12·x . . . . . . . matches the first choice

__

Additional comment

The equation of a hyperbola with vertices on the y-axis is often written as ...

  y²/a² -x²/b² = 1

We have written it in the above form to better show the scale factors applied to x and y.

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Answer:

[tex]\frac{4}{3}[/tex] or [tex]1\frac{1}{3}[/tex]

Step-by-step explanation:

[tex]1 + \frac{1}{3} \\= \frac{3}{3} + \frac{1}{1}\\= \frac{4}{3}[/tex]

(if the question is asking for the simplified form , it is [tex]1\frac{1}{3}[/tex]

Answer:

(4, -3)

Step-by-step explanation:

A vertex is just a point where 2 lines going in different directions meet. so the answer is (4, -3)

Answer:

[tex]11 - 2 {}^{3} [/tex]

[tex]11 - 8[/tex]

[tex]3[/tex]

The team gain 1490 yards by rushing during the season.

Football yardage is the number of yards that a team or player manages to move the ball forward toward their opponent's end zone.

According to the question,

A football teams offense T is found by adding the total passing yardage P to the total rushing yardage R,

i.e.  T=P+R

In 2012,

The teams total offense was 4559 yards.

The teams passing yardage was 3069 yards.

By using the formula

T=P+R

T = 4559 yards , P = 3069 yards

Put the values in the above equation;

4559 =  3069 - R

4559 - 3069 = R

R = 1490 yards

Therefore , 1490 yards the team gain rushing during the season.

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By the power and chain rules, taking derivatives on both sides with respect to [tex]t[/tex] gives

[tex]4x^3 \dfrac{dx}{dt} + 4y^3 \dfrac{dy}{dt} = 4\dfrac{dy}{dt}[/tex]

or

[tex]4x^3 \dfrac{dx}{dt} + (4y^3-4) \dfrac{dy}{dt} = 0[/tex]

The question is incomplete. The complete question is

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shown in the below-mentioned figure shows the results of the test.

Which class should get the reward, and why?

- The 2nd-period class should get the reward. They have the highest score, a perfect 100.

- The 2nd-period class should get the reward. They have a higher median.

- The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

- The 4th-period class should get the reward. Their lowest score is an outlier and should be thrown out.

The box plot shown in the question provides the information that
In 2nd period the minimum of the data is 77, first quartile is 78, median of the data is 89, third quartile is 92 and the maximum of the data is 100.
Hence, the mean of the results of 2nd period is

[tex]\frac{77+78+89+92+100}{5}=87.2[/tex]

In 2nd period the minimum of the data is 72, first quartile is 83, median of the data is 89, third quartile is 96 and the maximum of the data is 98.

Hence, the mean of the results of the 4th period is

[tex]\frac{72+83+89+96+98}{5}=87.6[/tex]

Hence, obviously, the 4th period is having more mean.

Moreover, we can observe that if more of the marks are on the higher side the average will automatically be more. Hence, as the first and the third quartiles are more on the 4th-period, so the average marks for the 4th period must be better.

So, just by observing the box plot, we can give the statement that  "The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class. "

Therefore, the 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

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Answer:

sum of all sides

Step-by-step explanation:

x1y2 + x3y3 + x1y1

The expression for the polynomial graphed will be y(x) = (x + 3)(x - 1 )(x - 4 ).

From the graph, the zeros of the polynomial of given graph are:

x = -3

x = 1

x = 4

Equate the above equations to zero

x + 3 = 0

x - 1 = 0

x - 4 = 0

Multiply the equations

(x + 3)(x - 1 )(x - 4 ) = 0

Express as a function gives;

y = (x + 3)(x - 1 )(x - 4 )

Hence, the factored form of the polynomial will be y = (x + 3)(x - 1 )(x - 4 ) .

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Answer:

200 sqin.

Step-by-step explanation:

10 x 10 + 10 x 10 = 100 + 100
                          = 200

The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.

Domain is the values of x in the function represented by y=f(x), for which y exists.

THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".

Now we assume the [tex]g(x)=x+2[/tex] and [tex]f(x)=\frac{1}{x-6}[/tex]

So here since g(x) is a polynomial function so it exists for all real x.

[tex]f(x)=\frac{1}{x-6}[/tex]  does not exists when [tex]x=6[/tex], so the domain of f(x) is given by all real x except 6.

Now,

[tex](fg)(x)=f(g(x))=f(x+2)=\frac{1}{(x+2)-6}=\frac{1}{x-4}[/tex]

So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.

But domain of both f(x) and g(x) consists of the value x=4.

Hence the statement is not TRUE universarily.

Thus the given statement about the composition of function is FALSE.

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28-8 is the discount price of each ticket, and the (28-8) is equal to (=) -----> 24(28-8

Step-by-step explanation:

that is the answer

The size of the colony after 10 days is  852.

The growth rate of the colony can be represented with an exponential equation with the form:

FV = P(1 + r)^n

Where:

55(1.73)^5 = 852

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Answer:

36

Step-by-step explanation:

40*.9

Answer:

68.5% seats filled

76% points earned

Step-by-step explanation:

Percentages are formed when one finds a ratio of two related quantities, usually comparing the first partial quantity to the amount that "should" be there.

[tex]\text{ratio}=\dfrac {\text{the "part"}}{\text{the whole}}[/tex]

For instance, if you have a pie, and you eat half of the pie, you're in effect imagining the original pie (the whole pie) cut into two equal pieces, and you ate one of them (the "part" of a pie that you ate).  To find the ratio of pie that you ate compared to the whole pie, we compare the part and the whole:

[tex]\text{ratio}=\dfrac {\text{the number of "parts" eaten}}{\text{the number of parts of the whole pie}}[/tex]

[tex]\text{ratio}=\dfrac {1}{2}[/tex]

If you had instead eaten three-quarters of the pie, you're in effect imagining the original pie cut into 4 equal pieces, and you ate 3 of them.

[tex]\text{ratio}=\dfrac {\text{the number of "parts" eaten}}{\text{the number of parts of the whole pie}}[/tex]

[tex]\text{ratio}=\dfrac {3}{4}[/tex]

There can be cases where the "part" is bigger than the whole.  Suppose that you are baking pies and we want to find the ratio of the pies baked to the number that were needed, the number of pies you baked is the "part", and the number of pies needed is the whole.  This could be thought of as the ratio of project completion.

If we need to bake 100 pies, and so far you have only baked 75, then our ratio is:

[tex]\text{ratio}=\dfrac {\text{the number of "parts" made}}{\text{the number of parts of the whole order}}[/tex]

[tex]\text{ratio}=\dfrac {75}{100}[/tex]

But, suppose you keep baking pies and later you have accidentally made more than the 100 total pies.... you've actually made 125 pies.  Even though it's the bigger number, the number of pies you baked is still the "part" (even though it's bigger), and the number of pies needed is the whole.

[tex]\text{ratio}=\dfrac {\text{the number of "parts" made}}{\text{the number of parts of the whole order}}[/tex]

[tex]\text{ratio}=\dfrac {125}{100}[/tex]

To find a percentage from a ratio, there are two small steps:

Going back to the pies:

When you ate half of the pie, your ratio of pie eaten was [tex]\frac{1}{2}[/tex]

Dividing the two numbers, the result is [tex]0.5[/tex]

Multiplying by 100 gives 50.  So, the percentage of pie that you ate (if you ate half of the pie) is 50%

When you ate three-quarters of the pie, the ratio was [tex]\frac{3}{4}[/tex]

Dividing the two numbers, the result is 0.75

Multiplying by 100 gives 75.  So, the percentage of pie that you ate (if you ate three-quarters of the pie) is 75%.

When you were making pies, and 100 pies were needed, but so far you'd only baked 75 pies, the ratio was [tex]\frac{75}{100}[/tex]

Dividing the two numbers, the result is 0.75

Multiplying by 100 gives 75.  So, the percentage of the project that you've completed at that point is 75%.

Later, when you had made 125 pies, but only 100 pies were needed, the ratio was [tex]\frac{125}{100}[/tex]

Dividing the two numbers, the result is 1.25

Multiplying by 100 gives 125%.  So, the percentage of pies you've made to complete the project at that point is 125%.... the number of pies that you've made is more than what you needed, so the baking project is more than 100% complete.

1.   27400 spectators n a 40000 seat stadium percentage.

Here, it seems that the question is asking what percentage of the stadium is full, so the whole is the 40000 seats available, and the "part" is the 27400 spectators that have come to fill those seats.

[tex]\text{ratio}=\dfrac {\text{the number of spectators filling seats}}{\text{the total number of seats in the stadium}}[/tex]

[tex]\text{ratio}=\dfrac {27400}{40000}[/tex]

Dividing gives 0.685.  Multiplying by 100 gives 68.5.  So, 68.5% of the seats have been filled.

2.   an archer scores 95 points out of a possible 125 points percentage

Here, it seems that the question is asking what percentage of the points possible were earned, so the whole is the 125 points possible, and the "part" is the 95 points that were earned.

[tex]\text{ratio}=\dfrac {\text{the number of points earned}}{\text{the total number of points possible}}[/tex]

[tex]\text{ratio}=\dfrac {95}{125}[/tex]

Dividing gives 0.76.  Multiplying by 100 gives 76.  So, 76% of points possible were earned.

Answer:

False, distributions having the same mean and standard deviation can have very different shape of distribution.

The scenario can be described using a piecewise function like:

f(x) = 1/x                  if  x < c.

f(x) = x                    if   x = c

f(x) = 1/(x + 73)       if x > c.

Remember that the limit only exists if the limit from left and the limit from the right give the same value.

Then, we can just define a piecewise function of the form:

f(x) = 1/x                  if  x < c.

f(x) = x                    if   x = c

f(x) = 1/(x + 73)       if x > c.

Clearly, this is not a continuous function.

Notice that:

[tex]f(c) = c.\\\\ \lim_{x \to c^{-}} f(x) = 1/c\\\\ \lim_{x \to c^{+}} f(x) = 1/(c + 73)[/tex]

So the limits from left and right are different, then:

[tex]\lim_{x \to c^{}} f(x)[/tex]

Does not exist.

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Based on the hints of distinguishing between radical and non-radical numbers, we have the following lists:

In this question we must order sets of numbers in ascending order. A hint consists in comparing non-radical numbers with the closest radical numbers whose results are integers.

In consequence, we obtain the following orders:

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Answer:

09

Step-by-step explanation:

07