Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $1,264 with a standard deviation of $150. What is a 90% confidence interval for the mean daily income this parking garage will generate

Answer:

a) The horizontal asymptote of [tex]C(t)[/tex] is [tex]c = 0[/tex].

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of [tex]C(t)[/tex] is the horizontal line, to which the function converges when [tex]t[/tex] diverges to the infinity. That is:

[tex]c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5}[/tex] (1)

[tex]c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)[/tex]

[tex]c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }[/tex]

[tex]c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }[/tex]

[tex]c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }[/tex]

[tex]c = \frac{0}{3+0}[/tex]

[tex]c = 0[/tex]

The horizontal asymptote of [tex]C(t)[/tex] is [tex]c = 0[/tex].

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.  

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

[tex]C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}[/tex]

[tex]C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}[/tex]

[tex]C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)[/tex]

Now we equalize the expression to zero:

[tex]\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0[/tex]

[tex]1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0[/tex]

[tex]\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0[/tex]

[tex]5-3\cdot t^{2} = 0[/tex]

[tex]t = \sqrt{\frac{5}{3} }\,h[/tex]

[tex]t \approx 1.291\,h[/tex]

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

[tex]C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}[/tex]

[tex]C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}[/tex]

[tex]C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}[/tex]

If we know that [tex]t \approx 1.291\,h[/tex], then the value of the second derivative is:

[tex]C''(1.291\,h) = -0.077[/tex]

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.

Answer:

x = 13

y = 8

Step-by-step explanation:

Answer/Step-by-step explanation:

Let's find the measure of the angles of ∆QNP.

∆QMN is am isosceles ∆, because it has two equal sides. Therefore, its base angles would be the same. Thus:

m<MNQ = ½(180 - 48) (one of the base angles of ∆QMN)

m<MNQ = ½(132) = 66°

Next, find m<QNP

m<QNP = 180° - m<MNQ (linear pair angles)

m<QNP = 180° - 66° (Substitution)

m<QNP = 114°

Next, find m<P

m<P = 180 - (m<QNP + m<PQN) (sum of ∆)

m<P = 180 - (114 + 33)

m<P = 180 - 147

m<P = 33°

Thus, in ∆QNP, there are two equal angles, namely, <P and <PQN.

An isosceles ∆ had two equal base angles. Therefore, ∆QNP must be an isosceles ∆.

An isosceles triangle is that the triangle must have two sides of equal length.

Triangle QNP is isosceles triangle because, QN = PN

In triangle QMN,  

        Since,  QM = QN

 So,  ∠QMN = ∠QNM

By property of triangle:

∠MQN + ∠QNM + ∠QMN = 180

   48 + 2 ∠QNM = 180

              ∠QNM = [tex]\frac{180-48}{2}[/tex] = 66  degree

  So, ∠QMN = ∠QNM = 66 degree

from figure,

    ∠QNM + ∠QNP = 180

                    ∠QNP = 180 - 66 = 114 degree.

In triangle QNP,  

              ∠QNP + ∠PQN + ∠QPN = 180

                        ∠QPN = 180 - 33 - 114 = 33 degree

Since,     ∠QNP = ∠QPN = 33 degree

Therefore, triangle QNP is isosceles triangle.

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Answer:

2.4m+1.8b−32.6

Step-by-step explanation:

Group like terms together

1. 2.4 m + \left( - 125 + 92.4 \right) + 1.8 b2.4m+(−125+92.4)+1.8b

2 . Add or subtract the numbers

2.4 m- 32.6 + 1.8 b2.4m−32.6+1.8b

3 . Rearrange the expression

2.4 m + 1.8 b- 32.62.4m+1.8b−32.6

Answer:

2.4m+1.8b-32.6

Step-by-step explanation:

2.4m- 125 + 1.8b+ 92.4​

2.4m+1.8b-32.6

Answer:

70.9,109.1,204.0,336.0

Step-by-step explanation:

The solution set for the equation 6sin² θ - 7 sinθ - 5 = 0 in the interval [0,360)is {θ∈ [0,360)∣θ = 230.48 ,309.521}

Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles.

The given equation is 6sin² θ - 7 sinθ = 5

Lets arrange in an order

6sin² θ - 7 sinθ - 5 = 0

a=6, b=-7 and c = -5

sinθ = 7±√49+120/12

Simplifying under the square root:

sinθ =7±17/12

Taking the positive solution first

sinθ =24/12

sinθ =2 (Not valid)

Now let's try the negative solution:

sinθ =7-17/12

sinθ =-10/12

sinθ =-5/6

θ= sin⁻¹(-5/6)

θ= −50.48⁰ ,−129.52⁰

we need to restrict our solution to the given interval [0,360), so we add 360 degrees to the negative angle:

θ=360−50.48 = 309.52 and 360 −129.52 =230.48

Therefore, the solution set for the equation 6sin² θ - 7 sinθ - 5 = 0 in the interval [0,360)is {θ∈ [0,360)∣θ = 230.48 ,309.521}

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Answer:

$26.25

Step-by-step explanation:

1.25 per day

21 days

1.25 × 21 = 26.25

Answer:

ans= 12-x=r1 coin...

Step-by-step explanation:

hope u can do..

Answer:

Domain=(-infinity,0)

Range: (-infinity,-1)

Step-by-step explanation:

Find the domain by finding where the equation is defined. The range is the set of values that correspond with the domain.

it's probably 17424 feet

Answer: In three days

Step-by-step explanation:

Answer:

yes, it is

Step-by-step explanation:

plug in the point (1, -1) (x,y) into the equation y = 9x - 10

-1 = 9(1)-10

-1 = 9-10

-1=-1

Answer:

125.3 yards

Step-by-step explanation:

He wants them to run the  so 110^2+60^2 =which is 125.3.

Hope this helps plz mark brainliest :D

Answer:

35.015

Step-by-step explanation:

see attached image

Answer:

m=-2

Step-by-step explanation:

Answer:

This graph has a slop of 0

Step-by-step explanation

see the included picture of the graph

1809.56ft³ <<<<<<<<<<<<<<<<Answer

Answer:

452.7 ft³

Step-by-step explanation:

sooooo first you find the radius of the base.

Radius=16/2=8

now you find the area of the base:

Area of a circle=2(pi)(radius)=50.3 ( rounded up to the nearest 10ths)

now to get the volume:

volume of a cylinder=Base*hight=50.3*9= 452.7 ft³

Answer:

im 90% sure its A

Step-by-step explanation:

Answer:

I got (4x+3)(3x+1)

Hope I help Please mark me BrainIIest please and thank you :)

Step-by-step explanation:

Answer:

line dancing, square dancing, etc

Step-by-step explanation:

Answer: square dancing, circle dancing, line dancing.

Hope this helps... Stay safe and have a great day/night!!! :D

Answer:

5 percent.

Step-by-step explanation:

Divide tax by the cost to see the percentage of the tax.

750/37.50=0.05

5 percent.

The required terms are 7, 3 and 0. The correct option is (c).

What is an arithmetic sequence?

An arithmetic sequence is a kind of sequence of numbers where the difference of any two consecutive terms are the same.

This difference is known as the common difference of the sequence.

The given sequence is as below,

33, 25, 18, 12, …….

The difference between the consecutive terms is given as,

D₁ = 25 - 33

    = -8

D₂ = 18 - 25

     = -7

D₃ = 12 - 18

    = -6

It is clear that the difference between the consecutive terms are in arithmetic progression.

Thus, the next differences can be found as,

D₄ = -6 + 1

     = -5

D₅ = -5 + 1

    = -4  

And, D₆ = -4 + 1

             = -3

The next three terms of the sequence can be written as,

12 - 5 =7, 7 - 4 = 3, 3 - 3 = 0.

Hence, the next three terms of the given sequence are 7, 3 and 0.

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Answer:

The correct answer would be D. It would not be a linear function because the rate of change is not constant.

Step-by-step explanation:

The rate of change is not constant since it is not adding by a same number each time.

Answer:

y intercept is 2 x =-3

Step-by-step explanation:

2x-3y=-6

subtract 2

-3y=-2×-6

divide by -3

y=2/3+2

×

0=2/3x+2

-2=2/3×

-3=x

Answer:

No

Step-by-step explanation:

This relation is not a function. This line has repeating x-values, it only has one so each y-value cannot have a unique x-value. For this reason, no vertical line will ever be a function.  

Answer:

The closest equivalent of the ratio of a to c will be:

a/c=1.680

Step-by-step explanation:

As 'a' is 63% of x

so

[tex]a=0.63x[/tex]

As 'c' is 3/8 of x

so

c=3/8

Thus, the closest equivalent of the ratio of a to c will be:

a/c = (0.63x)/(3/8x)

      = 1.680

Therefore, the closest equivalent of the ratio of a to c will be:

a/c=1.680

Answer:

(Choice B) Keeping A constant and decreasing B

(Choice C) Increasing A and keeping B constant

Step-by-step explanation:

Let us analyze these two options.

2√a - b /5

(Choice B) Keeping A constant and decreasing B

Let a = 100 and b= 30

2√100 - 30/5

= 2(10) - 6

= 20 -6= 14

But if A is kept constant and B is decreased to 20

2√100 - 20/5

= 2(10) - 4

= 20 -4= 16

We see the expression value has increased from 14 to 16.

Similarly analyzing Option C

(Choice C) Increasing A and keeping B constant

Let a = 100 and b= 30

2√100 - 30/5

= 2(10) - 6

= 20 -6= 14

Now increasing A to 121 and keeping B constant would give

Let a = 100 and b= 30

2√121 - 30/5

= 2(11) - 6

= 22 -6= 16

This also increases the expression's value

But the other two choice A and D do not give the same result.

Let a = 100 and b= 30

2√100 - 30/5

= 2(10) - 6

= 20 -6= 14

(Choice A)

Keeping A constant and increasing B

2√100 - 40/5

= 2(10) - 8

= 20 -8= 12

This decreases the value .

Also

(Choice D)

Decreasing A and keeping B constant

2√81 - 40/5

= 2(9) - 8

= 18 -8= 10

This also decreases value

The correct choices are B and C

Answer:

the acutal solution is B and D

Answer:

Tell whether the following equations represent a linear function.

10x + 2y = 4--linear equation✓

−x² + 3y = 19--quadratic equation

6x + 1/2y = 3--linear equation✓