H-20
———— = 0
-3

Solve for h

Answer:

[tex]m(d)=26\ d[/tex]  

where [tex]m[/tex] represents miles driven

and [tex]d[/tex] represents number of days driven.

Step-by-step explanation:

Given:

Total distance driven in miles = 70,000

Average distance driven each day = 26 miles

Taking average as unit rate of miles covered per day.

∴ we can say the car covers 26 miles per day.

Using unitary method to find miles driven in [tex]d[/tex] days.

In 1 day miles driven = 26

In [tex]d[/tex] days miles driven = [tex]26\times d =26\ d[/tex]

So, to find [tex]m[/tex] miles driven the expression can be written as:

[tex]m(d)=26\ d[/tex]  

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let [tex]D^c[/tex] be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that  one person in 10,000 people has a rare genetic disease.

So,[tex]P(D)=\frac{1}{10000}[/tex]

Only 0.4% of the people who don't have it test positive.

[tex]P(A|D^c)[/tex] = probability that a person is tested positive given he don't have a disease = 0.004

[tex]P(D^c)=1-\frac{1}{10000}[/tex]

Formula:[tex]P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}[/tex]

[tex]P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}[/tex]

P(D|A)=[tex]\frac{2470}{2471}[/tex]=0.9995

P(D|A)=[tex]0.9995[/tex]

A)The probability that someone who tests positive has the disease is 0.9995

(B)

[tex]P(D^c|A^c)[/tex]=probability that someone does not have disease given that he tests negative

[tex]P(A^c|D^c)[/tex]=probability that a person tests negative given that he does not have disease =1-0.004

=0.996

[tex]P(A^c|D)[/tex]=probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: [tex]P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}[/tex]

[tex]P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}[/tex]

[tex]P(D^c|A^c)=0.99999[/tex]

B)The probability that someone who tests negative does not have the disease is 0.99999

Answer:

The probability that their mean length is less than 11.2 inches is 0.5987

Step-by-step explanation:

Mean = 10.9 inches

Standard deviation = 1.2 inches

We are supposed to find If 25 items are chosen at random, what is the probability that their mean length is less than 11.2 inches

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

We are supposed to find P(x<11.2)

[tex]Z=\frac{11.2-10.9}{1.2}[/tex]

[tex]Z=0.25[/tex]

Refer the z table for p value

p value = 0.5987

Hence the probability that their mean length is less than 11.2 inches is 0.5987

Answer:

Step-by-step explanation:

Motrcycle: 2 hours  150/75 = 2 hours

Bus:  150/60= 2.5 hours

Truck:150/50= 3 hours

Bike:= 150/20= 7.5 hours

Answer: 0.0047

Step-by-step explanation:

Given : A manufacturer knows that their items have a normally distributed length, with a mean of 7.1 inches, and standard deviation of 1.7 inches.

i.e. [tex]\mu=7.1\text{ inches}[/tex]

[tex]\sigma=17\text{ inches}[/tex]

Sample size : n= 24

Let [tex]\overline{X}[/tex] be the sample mean.

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

Then, the probability that their mean length is less than 6.2 inches will be :-

[tex]P(\overline{x}<6.2)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{6.2-7.1}{\dfrac{1.7}{\sqrt{24}}})\\\\\approx P(z<-2.6)\\\\=1-P(z<2.6)\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=1-0.9953=0.0047\ \ \ [ \text{Using z-value table}][/tex]

hence,. the required probability = 0.0047

Answer:

0.0047

step-by-step explanation

Step-by-step explanation:

Let w be the work of stock the shelves and t be the time for Camille to the worl alone.

Dolores takes 7.2 hours.

[tex]\texttt{Rate of Dolores = }\frac{w}{7.2}[/tex]

[tex]\texttt{Rate of Camille = }\frac{w}{t}[/tex]

If they combine work is completed in 2.8 hours.

          That is

                       [tex]2.8=\frac{w}{\frac{w}{7.2}+\frac{w}{t}}\\\\2.8=\frac{7.2t}{t+7.2}\\\\2.8t+20.16=7.2t\\\\4.4t=20.16\\\\t=4.58hours[/tex]

   It takes 4.58 hours to stock the shelves if Camille were working alone

Answer:

The answer to your question is 13x² - 14x

Step-by-step explanation:

Process

1.- Divide the figure in to sections to get to rectangles (see the picture below)

2.- Get the area of each rectangle

3.- Add the areas

2.- Area of a rectangle = base x height

Rectangle 1

      Area 1 = (3x - 7) (x)

                 = 3x² - 7x

Rectangle 2

      Area 2 = (5x + 2)(2x)

                  = 10x² + 4x

3.- Total area

     Area = (3x² - 7x) + (10x² - 7x)

              =  13x² - 14x

The magnitude of the angular momentum of the particle relative to the point (0.9 m, 10 m) is [tex]{\text} 32.8 kg m^2/s[/tex].

Angular momentum is a physical quantity that measures the rotational motion of an object or system.

Given:

Mass = 4 kg,

velocity = 2 m/s

The following equation provides the particle's angular momentum (L):

L = mvr

where:

m = mass

v = velocity of the particle

r = perpendicular distance

To find the magnitude of the angular momentum relative to the point point [tex](x_0, y_0)[/tex], where [tex]x_0[/tex] = 0.9 m and [tex]y_0[/tex] = 10 m.

To find the perpendicular distance (r), use the distance formula:

[tex]r = \sqrt{((x - x0)^2 + (y - y0)^2)[/tex]

Substituting the values [tex]x_0[/tex] = 0.9 m and [tex]y_0[/tex] = 10 m in above formula

[tex]r = \sqrt{((0 - 0.9)^2 + (6-10)^2)[/tex]

 = [tex]\sqrt{((-0.9)^2 + (-4)^2)[/tex]

 = √(0.81 + 16)

 = √16.81

 = 4.1 m

Now, the angular momentum (L) using the formula:

L = mvr

L = 4 kg x 2 m/s x 4.1 m

L = 32.8 kg

As a result, the particle's angular momentum is  [tex]{\text} 32.8 kg m^2/s[/tex].

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The magnitude of the angular momentum relative to a point (x0, y0) depends on the moment of inertia and angular velocity of the particle. However, in this case, the angular velocity is undefined, so the magnitude of the angular momentum is also undefined.

The angular momentum of a particle can be calculated by multiplying its moment of inertia with its angular velocity. In this case, the particle has a mass of 4 kg and moves with a constant speed of 2 m/s in the positive x-direction. To find the magnitude of its angular momentum relative to the point (x0, y0), we need to calculate the moment of inertia and angular velocity. Since the particle is moving in the xy-plane, we can calculate the distance of the particle from the point (x0, y0) and use it to find the angular momentum. The magnitude of the angular momentum can be calculated by dividing the cross product of the position vector and linear momentum with the mass of the particle.

First, let's calculate the moment of inertia (I) of the particle. The moment of inertia can be calculated using the formula I = mr², where m is the mass of the particle and r is the distance of the particle from the axis of rotation. In this case, the particle is moving in the xy-plane, so the distance of the particle from the point (x0, y0) can be calculated using the distance formula: d = sqrt((x-x0)² + (y-y0)²). Substituting the values, we have d = sqrt((0-0.9)² + (6-10)²) = sqrt(13.21) = 3.63 m. The moment of inertia can be calculated as I = 4 kg * (3.63 m)² = 52.60 kg*m².

Next, let's calculate the angular velocity (ω) of the particle. The angular velocity can be calculated using the formula ω = v/r, where v is the linear velocity of the particle and r is the distance of the particle from the axis of rotation. In this case, the particle has a constant speed of 2 m/s in the positive x-direction along y = 6 m, so the distance of the particle from the axis of rotation is the distance from the point (0, 6). Substituting the values, we have r = sqrt((0-0)² + (6-6)²) = sqrt(0) = 0 m. The angular velocity can be calculated as ω = 2 m/s / 0 m = undefined. Since the angular velocity is undefined, the magnitude of the angular momentum relative to the point (x0, y0) is also undefined.

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Answer:

Part 1) An expression for the x-coordinate of T is (a+2)

Part 2) The value of x=39 ft (see the explanation)

Step-by-step explanation:

Part 1)

step 1

we know that

The rule of the reflection of a point across the x-axis is equal to

[tex](x,y) -----> (x,-y)[/tex]

Apply the rule of the reflection across the x-axis to the Q coordinates

Q (a,b) ----------> Q'(a,-b)

step 2

The translation is 2 units to the right

so

The rule of the translation is

(x,y) ----> (x+2,y)

Apply the rule of the translation to the Q' coordinates

Q'(a,-b) -----> T(a+2,-b)

therefore

An expression for the x-coordinate of T is (a+2)

Part 2)

we know that

A reflection is a rigid transformation, the image is the same size and shape as the pre-image

In this problem the floor plan house A and the floor house B have the same size and shape

That means that its corresponding sides and corresponding angles are congruent

therefore

The value of x=39 ft

For this case we have the following equation:

[tex]I = \frac {nE} {nr + R}[/tex]

We must clear the variable "n", for them we follow the steps below:

We multiply by [tex]nr + R[/tex] on both sides of the equation:

[tex]I (nr + R) = nE[/tex]

We apply distributive property on the left side of the equation:

[tex]Inr + IR = nE[/tex]

Subtracting [tex]nE[/tex] from both sides of the equation:

[tex]Inr-nE + IR = 0[/tex]

Subtracting IR from both sides of the equation:

[tex]Inr-nE = -IR[/tex]

We take common factor n from the left side of the equation:

[tex]n (Ir-E) = - IR[/tex]

We divide between Ir-E on both sides of the equation:

[tex]n = - \frac {IR} {Ir-E}[/tex]

Answer:

[tex]n = - \frac {IR} {Ir-E}[/tex]

Answer:

2

Step-by-step explanation:

y=a sin x

amplitude=|a|

a=-2

amplitude=2

Answer:

d=1.416.525 mile

Step-by-step explanation:

V1=5151m/h, t1=t, V2=7575m/h, t2=t-88h

d1=d2 Because it is same distance; V1=d1/t and V2=d2/(t-88) but d1=d2

d=V1t=V2(t-88) → 5151t=7575(t-88) → 5151t=7575t-666.600 → 7575t-5151t=666.600 → 2424t=666.600 → t=666.600/2424 → t=275h so

[tex]d=5151\frac{mile}{h}.275h =  1.416.525mile[/tex]

Answer:

Marco will use [tex]1[/tex] red tile, [tex]2[/tex] blue tiles and [tex]1\frac{2}{3}[/tex] yellow tiles in each stone.

Step-by-step explanation:

Given:

Number of red tiles = 45

Number of blue tiles = 90

Number of yellow tiles = 75

Greatest number of stones that Marco can make = 45

To determine how many each color tile Marco will use in each stone.

Solution:

In order to determine the number of each color tile in each stone we need to divide total number of a particular tile by total number of stones. By doing this we can get the exact number of that color tile used in each stone.

Number of red tile in each stone

⇒ [tex]\frac{\textrm{Total number of red tiles}}{\textrm{Total number of stones}}[/tex]

⇒ [tex]\frac{45}{45}=1[/tex]

Number of blue tiles in each stone

⇒ [tex]\frac{\textrm{Total number of blue tiles}}{\textrm{Total number of stones}}[/tex]

⇒ [tex]\frac{90}{45}=2[/tex]

Number of yellow tile in each stone

⇒ [tex]\frac{\textrm{Total number of yellow tiles}}{\textrm{Total number of stones}}[/tex]

⇒ [tex]\frac{75}{45}[/tex]

⇒ [tex]\frac{5}{3}[/tex]   [Reducing to simpler fraction by dividing both numbers by their GCF=15]

⇒ [tex]1\frac{2}{3}[/tex]  [Converting improper fraction to mixed number]

∴ Marco will use [tex]1[/tex] red tile, [tex]2[/tex] blue tiles and [tex]1\frac{2}{3}[/tex] yellow tiles in each stone.

Answer:

Marco will use 3 red, 6 blue and 5 yellow tiles on each stones

Explanation:

Given, each stone must have same number of each colour tile

Then, calculating the highest common factors of the number HCF (45, 90, 75)

Factors of 45 = 3 × 3  ×  5

Factors of 90 = 2  × 3 × 3  × 5

Factors of 75 = 3 × 5  ×  5

Highest Common Factors (HCF) = 3 × 5 = 15

Dividing all three numbers by 15, we get

Red Tiles =[tex]\frac{45}{15}[/tex] = 3

Blue Tiles = [tex]\frac{90}{15}[/tex] = 6

Yellow Tiles = [tex]\frac{75}{15}[/tex] = 5

Therefore, Marco will use 3 red, 6 blue and 5 yellow tiles on each stones

Answer:

  14 feet

Step-by-step explanation:

There are 3 feet in 1 yard, so 36 feet in 12 yards. The remaining ribbon will be the original amount less the amount used.

  36 - 22 = 14 . . . . feet remaining

After decorating the blankets, Olga will have 14 feet of ribbon remaining. The conversion from yards to feet and subtraction calculates this remaining amount accurately.

Calculating Remaining Ribbon

To determine how much ribbon Olga has left after decorating the blankets, we need to perform a couple of conversions and a subtraction.

First, let's convert the total ribbon from yards to feet:

→ 1 yard = 3 feet

→ 12 yards = 12 * 3

                 = 36 feet

Next, Olga uses 22 feet of ribbon to decorate the blankets:

→ Total ribbon in feet: 36 feet

→ Ribbon used: 22 feet

Now, subtract the amount used from the total:

→ Remaining ribbon = 36 feet - 22 feet

                                 = 14 feet

Olga will have 14 feet of ribbon remaining.

Answer:

3.3 in other words between 2-5

Step-by-step explanation:

33% of 100 is 33 so

33% of 10 is 3.3

Answer:

i would think that its c

let me know if its wrong

Answer:

c

Step-by-step explanation:

i know

Answer:

Step-by-step explanation:

what?

Answer:

Step-by-step explanation:

The relationships between the radius of a circle and its circumference and area are ...

  C = 2πr

  A = πr²

The relationships between the side length of a square and its perimeter and area are ...

  P = 4s

  A = s²

So, the length of wire will be ...

  w = C + P

  w = 2πr + 4s

subject to the constraint that the sum of areas is 64 cm²:

  πr² + s² = 64

___

Using the method of Lagrange multipliers to find the extremes of wire length, we want to set the partial derivatives of the Lagrangian (L) to zero.

  L = 2πr + 4s + λ(πr² +s² -64)

  ∂L/∂r = 0 = 2π +2πλr . . . . . . [eq1]

  ∂L/∂s = 0 = 4 +2λs . . . . . . . . [eq2]

  ∂L/∂λ = 0 = πr² +s² -64 . . . . [eq3]

__

Solving for λ, we find ...

  0 = 1 +λr . . . . divide [eq1 by 2π

  λ = -1/r . . . . . . subtract 1, divide by r

Substituting into [eq2], we get ...

  0 = 4 + 2(-1/r)s

  s/r = 2 . . . . . . . . . .add 2s/r and divide by 2

This tells us the maximum wire length is that which makes the circle diameter equal to the side of the square.

Substituting the relation s=2r into the area constraint, we find ...

  πr² +(2r)² = 64

  r = √(64/(π+4)) = 8/√(π+4) ≈ 2.99359 . . . . cm

and the maximum wire length is ...

  2πr +4(2r) = 2r(4+π) = 16√(4+π) ≈ 42.758 . . . cm

_____

The minimum wire length will be required when the entire area is enclosed by the circle. In that case, ...

  πr² = 64

  r = √(64/π)

  C = 2πr = 2π√(64/π) = 16√π ≈ 28.359 . . . cm

_____

Comment on the solution method

The method of Lagrange multipliers is not needed to solve this problem. The alternative is to write the length expression in terms of one of the figure dimensions, then differentiate with respect to that:

  w = 2πr + 4√(64-πr²)

  dw/dr = 2π -4πr/√(64-πr²) = 0

  64 -πr² = 4r²

  r = √(64/(π+4)) . . . . same as above

_____

Comment on the graph

The attached graph shows the relationship between perimeter and circumference for a constant area. The green curve shows the sum of perimeter and circumference, the wire length. The points marked are the ones at the minimum and maximum wire length.

The minimum length of the wire can be found by setting up a function to represent the total length of the wire and using calculus to minimize it. The maximum length of the wire is undefined because the length of the wire can increase indefinitely as the radius of the circle decreases.

To solve this problem, we use the formulas for the perimeters of a square and a circle, and the fact that the sum of their areas should equal 64 cm2. The perimeter of a square is 4s and the circumference of a circle is [tex]2\pi r[/tex], where s and r represent the side length of the square and the radius of the circle, respectively. The area of a square is s2 and the area of a circle is πr2.

The total length of the wire is the sum of the perimeter of the square and the circumference of the circle. The total area enclosed by the wire, according to the problem, should be 64 cm2.

To find the minimum length of wire needed, we can use calculus to minimize the function representing the length of the wire. The maximum length of the wire is undefined because as the radius of the circle approaches zero, the side length of the square and therefore the length of the wire can increase indefinitely.

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Answer:

  yes

Step-by-step explanation:

The graph passes the vertical line test, so is the graph of a function (yes). Each input value has exactly one output value.

Answer:

61°C

Step-by-step explanation:

Newton's Law  of cooling gives the temperature -time relationship has:

T (t)  = 20 + 75 е⁻(t/50)-------------------------------------------------------- (1)

where Time  is in  minutes (min) & Temperature in degree Celsius (°C)

During the first half hour, t = 30 mins

Substituting into (1)

T = 20 + 75  е⁻(30/50)

  = 20 + 75(0.5488)

  =  20 + 41.16

  =  61.16°C

  ≈    61°C

Answer:

You didn't write the expression

Step-by-step explanation:

Answer:

Y - intercept of equation of-line A is y = 1.

Y - intercept of equation of-line B is y = -2.

Step-by-step explanation:

Given:

For Iine A:

[tex]y+ 1 =\frac{1}{5}\times (x+10)[/tex]

For line B:

x = -2  then y = 2

x = -1  then y = 0

x = 0  then y = -2

x = 1  then y = -4

To Find:

Y- intercepts of Line A and Line B.

Solution:

Intercepts: Where the line cut X axis called X- intercept and where cut Y axis is called Y- intercept.

Y-intercept mean x coordinate will be 0

Therefore Put x = 0 in Line A we get

[tex]y+ 1 =\frac{1}{5}\times (0+10)\\y+1=\frac{10}{5}\\ y+1=2\\y= 2-1\\y=1[/tex]

Y - intercept of equation of-line A is y = 1.

For line B

See where x coordinate is 0 ,Therefore we have,

y = -2

Y - intercept of equation of-line B is y = -2.

1.what is the length of the segment joining 3,6 and -2,-6 : 13 units

2.what is the center of the circle (x+6)^2+(y-8)^2=144 => (-6,8)

3.what is the slope of the line 3y+2x-6=0=> -2/3

Step-by-step explanation:

1.what is the length of the segment joining (3,6) and (-2,-6)?

Let

(x1,y1) = (3,6)

(x2,y2) = (-2,-6)

The length of a segment is given by:

[tex]d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Putting\ values\\d  = \sqrt{(-2-3)^2+(-6-6)^2}\\d = \sqrt{(-5)^2+(-12)^2}\\= \sqrt{25+144}\\= \sqrt{169}\\=13\ units[/tex]

2.what is the center of the circle (x+6)^2+(y-8)^2=144

The equation of circle is given by:

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

Here, h and k are the coordinates of centre of circle

x - h = x+6

-h = 6

h = -6

y - 8 = y - k

-8 = - k

k = 8

So,

The center of circle is: (-6,8)

3.what is the slope of the line 3y+2x-6=0

We have to convert the equation in slope-intercept form to find the slope

Slope-intercept form is:

y = mx+b

Now,

[tex]3y+2x-6=0\\3y+2x = 6\\3y = -2x+6[/tex]

Dividing both sides by 3

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

In slope-intercept form, the co-efficient of x is the slope of the line so

m = -2/3

Keywords: Coordinate geometry, Slope

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Answer:

Does not exceed 2% in both cases.

Step-by-step explanation:

Given that a  manufacturer of interocular lenses will qualify a new grinding machine if there is evidence that the percentage of polished lenses that contain surface defects does not exceed 2%.

Sample proportion = [tex]\frac{6}{250} \\=0.024[/tex]

Create hypothesses as

[tex]H_0: p = 0.02\\H_a : p >0.02[/tex]

(Right tailed test at 5% significance level)

P difference = 0.004

Std error = 0.0089

test statistic Z = p diff/std error = 0.4518

p value = 0.326

Since p >alpha, we accept nullhypothesis

b) For confidence interval 97% we have

Margin of error = 2.17* std error = 0.0192

Confidence interval

= [tex](0.024-0.0191, 0.024+0.0191)\\= (0.0049, 0.0431)\\[/tex]

Since 2% = 0.02 lies within this interval we accept null hypothesis.

Does not exceed 2%

Since

Final answer:

To determine whether the new grinding machine can be qualified, we need to test the hypothesis that the percentage of polished lenses with surface defects does not exceed 2%.

Explanation:

To determine whether the new grinding machine can be qualified, we need to test the hypothesis that the percentage of polished lenses with surface defects does not exceed 2%. The null hypothesis (H₀) is that the proportion of defective lenses is equal to or less than 2%, while the alternative hypothesis (Ha) is that the proportion of defective lenses is greater than 2%. We can use a one-sample proportion test to analyze the data.

(a) The hypotheses are:

H₀: p ≤ 0.02 (proportion of defective lenses)

Ha: p > 0.02 (proportion of defective lenses)

With a significance level of α = 0.05, we can calculate the p-value from the sample data. Using a normal approximation to the binomial distribution, we find that the p-value is 0.0165.

(b) The question in part (a) can also be answered using a confidence interval. We can calculate a confidence interval for the proportion of defective lenses and see if it includes or excludes the value of 0.02. If the confidence interval includes 0.02, it suggests that the machine can be qualified. If the confidence interval does not include 0.02, it suggests that the machine cannot be qualified.

Answer:

(3) ∠BCA ≅ ∠DAC

Step-by-step explanation:

BC and AD are parallel.  AC is a transversal line passing through both lines.  That means ∠BCA and ∠DAC are alternate interior angles.  Therefore, they are congruent.

The formula for the value of nth term is [tex]a_{n}[/tex] = 3n + 1

Step-by-step explanation:

The formula of the nth term in the arithmetic sequence is

[tex]a_{n}=a+(n-1)d[/tex] , where

∵ The first term of an arithmetic sequence is equal to four

∴ a = 4

∵ The common difference is equal to three

∴ d = 3

- Substitute these values in the rule of the nth term

∵ [tex]a_{n}=a+(n-1)d[/tex]

∴ [tex]a_{n}=4+(n-1)3[/tex]

- Simplify it

∴ [tex]a_{n}=4+3n-3[/tex]

∴ [tex]a_{n}=1+3n[/tex]

The formula for the value of nth term is [tex]a_{n}[/tex] = 3n + 1

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Answer:

a. $104.17 monthly interest.

b. 120 monthly payments.

c. Total interest of $12,500.

Step-by-step explanation:

a.

I = Prt

I = (20000 x 0.0625 x 1) = 1250 annually

for monthly Interest payment divide the answer by 12;

1250/12 = $104.17 monthly

b.

12 x 10 = 120 monthly payments

c.

I = Prt

I = $20,000 x 0.0625 x 10

I = $12,500

The interest rate for the $4000 investment is approximately 2.1% and for the $20000 investment, it is approximately 2.7%.

This problem can be solved using the concept of simple interest. Let's denote the interest rate of the $4000 investment as r (in decimal form) and the interest rate of the $20000 investment would be r+0.006. Now, we can set up our equations based on the information given:

1. For $4000 investment: Interest = 4000 * r

2. For $20000 investment: Interest = 20000 * (r + 0.006)

It is also given that the interest earned from the $20000 investment is $1320 more than the $4000 investment. Therefore, we can set up a third equation as:

20000 * (r + 0.006) - 4000 * r = 1320

By solving this equation, we find that r (corresponding to the $4000 investment) is approximately 0.021 or 2.1% and therefore, the interest rate for the $20000 investment is roughly 2.7% (2.1% + 0.6%).

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Answer:

The dimensions of the pen that minimize the cost of fencing are:

[tex]x \approx 12.25 \:ft[/tex]

[tex]y \approx 8.17 \:ft[/tex]

Step-by-step explanation:

Let [tex]x[/tex] be the width and [tex]y[/tex] the length of the rectangular pen.

We know that the area of this rectangle is going to be [tex]x\cdot y[/tex].The problem tells us that the area is 100 feet, so we get the constraint equation:

[tex]x\cdot y=100[/tex]

The quantity we want to optimize is going to be the cost to make our fence. If we have chain-link on three sides of the pen, say one side of length [tex]y[/tex] and both sides of length [tex]x[/tex], the cost for these sides will be

[tex]10(y+2x)[/tex]

and the remaining side will be fence and hence have cost

[tex]20y[/tex]

Thus we have the objective equation:

[tex]C=10(y+2x)+20y\\C=10y+20x+20y\\C=30y+20x[/tex]

We can solve the constraint equation for one of the variables to get:

[tex]x\cdot y=100\\y=\frac{100}{x}[/tex]

Thus, we get the cost equation in terms of one variable:

[tex]C=30(\frac{100}{x})+20x\\C=\frac{3000}{x}+20x[/tex]

We want to find the dimensions that minimize the cost of the pen, for this reason, we take the derivative of the cost equation and set it equal to zero.

[tex]\frac{d}{dx} C=\frac{d}{dx} (\frac{3000}{x}+20x)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\C'(x)=\frac{d}{dx}\left(\frac{3000}{x}\right)+\frac{d}{dx}\left(20x\right)\\\\C'(x)=-\frac{3000}{x^2}+20[/tex]

[tex]C'(x)=-\frac{3000}{x^2}+20=0\\\\-\frac{3000}{x^2}x^2+20x^2=0\cdot \:x^2\\-3000+20x^2=0\\-3000+20x^2+3000=0+3000\\20x^2=3000\\\frac{20x^2}{20}=\frac{3000}{20}\\x^2=150\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{150},\:x=-\sqrt{150}[/tex]

Because length must always be zero or positive we take [tex]x=\sqrt{150}[/tex] as only value for the width.

To check that this is indeed a value of [tex]x[/tex] that gives us a minimum, we need to take the second derivative of our cost function.

[tex]\frac{d}{dx} C'(x)=\frac{d}{dx} (-\frac{3000}{x^2}+20)\\\\C''(x)=-\frac{d}{dx}\left(\frac{3000}{x^2}\right)+\frac{d}{dx}\left(20\right)\\\\C''(x)=\frac{6000}{x^3}[/tex]

Because [tex]C''(\sqrt{150})=\frac{6000}{\left(\sqrt{150}\right)^3}=\frac{4\sqrt{6}}{3}[/tex] is greater than zero, [tex]x=\sqrt{150}[/tex] is a minimum.

Now, we need values of both x and y, thus as [tex]y=\frac{100}{x}[/tex], we get

[tex]x=\sqrt{150}=5\sqrt{6}=12.25[/tex]

[tex]y=\frac{100}{\sqrt{150}}=\frac{10\sqrt{6}}{3}\approx 8.17[/tex]

The dimensions of the pen that minimize the cost of fencing are:

[tex]x \approx 12.25 \:ft[/tex]

[tex]y \approx 8.17 \:ft[/tex]

It can be done in 50388 ways

Step-by-step explanation:

When the selection has to be made without order, combinations are used.

The formula for combination is:

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

Here

Total books = n =19

Books to be chosen = r = 7

Putting the values

[tex]C(19,7) = \frac{19!}{7!(19-7)!}\\\\=\frac{19!}{7!12!}\\\\=50388\ ways[/tex]

It can be done in 50388 ways

Keywords: Combination, selection

Learn more about combinations at:

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