Is this equation linear or nonlinear?
y =x/2​

The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.

Answer:

The better deal would be simple interest rate of 3%

Step-by-step explanation:

In order to calculate which bank would be the better deal If Trsitam decides to deposit $7,000 for 4 years, we would have to make the following calculation:

simple interest rate of 3%.

Therefore, I= P*r*t

=$7,000*3%*4

I=$840

FV= $7,000+$840

FV=7,840

compound interest rate of 2.5%

Therefore, FV=PV(1+r)∧n

FV=$7,000(1+0.25)∧4

FV=$17,089

The better deal would be simple interest rate of 3%

Answer:

(Going from left to right)

Box #1=3

Box #2=5

Box #3=7

Box #4=2

Step-by-step explanation:

For Box #4, there is nothing for the 2 to subtract from so it just goes down

For Box #3, it has to be 7, because nothing can be subtracted from 1 to get 3, so you would have to bring a 1 from the 4 to the left to make the 1 to a 10. 10-7=3

For Box #2 and 1, 3(we changed it in the last step) -9 = a negative number so we have to bring a 1 from the number to the left. This is a hard step but what you have to do it look at the bottom number, which is a 2, so that number had to be a 3 because 3-1=2. 4 becomes 14, and 14-9=5

Hope this helps, if it did, please consider giving me brainliest, it will help me a lot. If you have any questions, feel free to ask.

Have a good day! :)

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation of the answer:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

▬▬▬▬▬▬▬▬▬▬▬▬

Answer:

3/5

Step-by-step explanation:

5x= 3

x= 3/5

hope you understand the answer

stay at home stay safe

keep rocking

pls mark me as BRAINLIEST

Answer:

Product of given question is 6√5

Step-by-step explanation:

Given:

3√2(√10)

Find:

Product.

Computation:

⇒ 3√2(√10)

⇒ 3√20

⇒ 3√4×5

⇒ 3√2×2×5

⇒ 3×2√5

⇒ 6√5

Product of given question is 6√5

Answer: 3rd Degree

Step-by-step explanation:

Answer:

(A) [tex]\text{The slope of OA is }\dfrac{4}{3}[/tex]

(C) It’s equation is 4x-3y=0.

Step-by-step explanation:

Point O is at (0,0)

Point A is at (3,4)

[tex]\text{Slope of OA}=\dfrac{4-0}{3-0} \\m=\dfrac{4}{3}[/tex]

The equation of a straight line is in the form: y=mx+b

The y-intercept of the line OA=0

Therefore, we have:

[tex]y=\dfrac{4}{3}x+0\\3y=4x+0\\$Subtract 3y from both sides$\\4x-3y=0[/tex]

The equation of the line is: 4x-3y=0

Answer:

can you please simplify that in English?

The volume is given by the integral,

[tex]\displaystyle\int_0^{2\pi}\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\int_0^4\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta[/tex]

That is, [tex]\rho[/tex] ranges from the origin to the sphere of radius 4. The range for [tex]\varphi[/tex] starts at the intersection of the cone [tex]z=x^2+y^2[/tex] with the sphere [tex]x^2+y^2+z^2=16[/tex], which gives

[tex]z+z^2=16\implies z^2+z-16=0\implies z=\dfrac{\sqrt{65}-1}2[/tex]

and

[tex]z=4\cos\varphi\implies\varphi=\cos^{-1}\left(\dfrac{\sqrt{65}-1}8\right)[/tex]

and extends to the x-y plane where [tex]\varphi=\frac\pi2[/tex]. The range for [tex]\theta[/tex] is self-evident.

The volume is then

[tex]V=\displaystyle\int_0^{2\pi}\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\int_0^4\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta[/tex]

[tex]V=\displaystyle\left(\int_0^{2\pi}\mathrm d\theta\right)\left(\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\sin\varphi\,\mathrm d\varphi\right)\left(\int_0^4\rho^2\,\mathrm d\rho\right)[/tex]

[tex]V=2\pi\left(\dfrac{\sqrt{65}-1}8\right)\left(\dfrac{64}3\right)=\boxed{\dfrac{16\pi(9-\sqrt{65})}3}[/tex]

A sphere is a three-dimensional object with a round form. The volume of the sphere is [16π(9-√65)]/3 unit³.

A sphere is a three-dimensional object with a round form. A sphere, unlike other three-dimensional shapes, has no vertices or edges. Its centre is equidistant from all places on its surface. In other words, the distance between the sphere's centre and any point on its surface is the same.

We know that the volume of the given sphere can be given by the integral,

[tex]{\rm Volume} = \int^{2\pi}_0\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} \int_0^4\rho^2sin\varphi\ d\rho\ d\varphi\ d\theta[/tex]

where ρ ranges from the origin of the plot to the sphere of radius 4 while the range of φ starts at the intersection of the cone z=x²+y² with the sphere x²+y²+z²=16.

Now, the value of z and φ can be written as,

[tex]x^2+y^2+z^2 = 16\\\\(x^2+y^2)+z^2 = 16\\\\z+z^2 = 16\\\\z^2+z-16=0 \implies z=\dfrac{\sqrt{65}-1}{2}[/tex]

And

[tex]z =4\ cos\ \varphi \implies \varphi =cos^{-1}(\dfrac{\sqrt{65}-1}{8})[/tex]

Further, the volume of the sphere can be written as,

[tex]{\rm Volume} = \int^{2\pi}_0\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} \int_0^4\rho^2sin\varphi\ d\rho\ d\varphi\ d\theta\\\\\\{\rm Volume} = (\int^{2\pi}_0\ d\theta)(\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} sin\varphi\ d\varphi)(\int_0^4\rho^2 d\rho)\\\\\\V = 2\pi(\dfrac{\sqrt{65}-1}{8})(\dfrac{64}{3}) = \dfrac{16\pi(9-\sqrt{65})}{3}[/tex]

Hence, the volume of the sphere is [tex]\dfrac{16\pi(9-\sqrt{65})}{3}[/tex].

Learn more about Sphere:

https://brainly.com/question/11374994

Answer:

Angle 2= 45

Angle 3= 45

Angle 4= 135

Angle 5= 135

Angle 6= 45

Angle 7= 45

Angle 8= 135

Answer:

4x² + 30x - 2 = 0

Step-by-step explanation:

Given:

Area = 58 square feet

Width = 7 feet

Length = 8 feet

Since the area is 58, writing the equation, we have:

(8 + 2x)(7 + 2x) = 58

Now expand the equation:

56 + 16x + 14x + 4x² = 58

56 + 30x + 4x² = 58

Collect like terms:

30x + 4x² + 56 - 58 = 0

30x + 4x² - 2 = 0

Rearrange the equation to a proper quadratic equation:

4x² + 30x - 2 = 0

The quadratic equation that can be used to determine the thickness of the frame, x is 4x² + 30x - 2 = 0

Answer:

m<1 = 30

Step-by-step explanation:

To find m<1, we can do 180 - 75 - 75, which will give us 30 degrees, so m<1 = 30

Answer:

C.I =  0.7608   ≤ p ≤   0.8392

Step-by-step explanation:

Given that:

Let consider a  random sample n = 400 candidates where  320 residents indicated that they voted for Obama

probability [tex]\hat p = \dfrac{320}{400}[/tex]

= 0.8

Level of significance ∝ = 100 -95%

= 5%

= 0.05

The objective is to  develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.

The confidence internal can be computed as:

[tex]=\hat p \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }[/tex]

where;

[tex]Z_{0.05/2}[/tex] = [tex]Z_{0.025}[/tex] = 1.960

SO;

[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }[/tex]

[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }[/tex]

[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }[/tex]

[tex]=0.8 \pm 1.960 \sqrt{4 \times 10^{-4}}[/tex]

[tex]=0.8 \pm 1.960 \times 0.02}[/tex]

[tex]=0.8 \pm 0.0392[/tex]

= 0.8 - 0.0392     OR   0.8 + 0.0392  

= 0.7608    OR    0.8392

Thus; C.I =  0.7608   ≤ p ≤   0.8392

Answer:

[tex]\boxed{Mean = 34.33}[/tex]

Step-by-step explanation:

Mean = Sum of Observations / No. Of Observations

Mean = (24+18+37+82+17+26)/6

Mean = 206 / 6

Mean = 34.33

Answer:

The Man needs to run at 9 mph

Step-by-step explanation:

Let M stand for the man's speed in mph.  When the man  

runs toward point A, the relative speed of the train with respect  

to the man is the train's speed plus the man's speed (45 + M).  

When he runs toward point B, the relative speed of the train is the  

train's speed minus the man's speed (45 - M).

When he runs toward the train the distance he covers is 2 units.  

When he runs in the direction of the train the distance he covers  

is 3 units. We can now write that the ratio of the relative speed  

of the train when he is running toward point A to the relative speed  

of the train when he is running toward point B, is equal to the  

inverse ratio of the two distance units or

              (45 + M)          3

              -----------  =      ---

              (45 - M)          2

          90+2 M=135-3 M

⇒5 M = 45

⇒ M = 9 mph

The Man needs to run at 9 mph

Answer: 9 mph

Step-by-step explanation:

Given that a man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph .

If the man tend to run in the forward direction, he will cover another 2/5 before the train reaches his initial position. The distance covered by the man will be 2/5 + 2/5 = 4/5

The remaining distance = 1 - 4/5 = 1/5

If the man can run at a constant rate in either direction to get off the bridge just in time before the train hits him, the time it will take the man will be

Speed = distance/time

Time = 1/5d ÷ speed

The time it will take the train to cover the entire distance d will be

Time = d ÷ 45

Equate the two time

1/5d ÷ speed = d ÷ 45

Speed = d/5 × 45/d

Speed = 9 mph

Answer:

y=4x

Step-by-step explanation:

Add 4x to both sides to get y=mx+b

0 is y-intercept.

4x is the slope.

Answer:

x=18

Step-by-step explanation:

Answer:

x = 18

Step-by-step explanation:

Since 1/3x = 6,

x = 6 x 3

Thus, x = 18

Answer:

Probability that it'll not rain during the next three days = 0.064

Step-by-step explanation:

Given

Let:

P(R) represent the probability that it'll rain each day

P(R') represent the probability that it'll not

[tex]P(R) = 60\%[/tex]

Required

Probability that it'll not rain during the next three days

From concept of probability;

[tex]P(R) + P(R') = 1[/tex]

Substitute 60% for P(R)

[tex]60\% + P(R') = 1[/tex]

Subtract 60% from both sides

[tex]60\% - 60\% + P(R') = 1 - 60\%[/tex]

[tex]P(R') = 1 - 60\%[/tex]

Convert % to decimal

[tex]P(R') = 1 - 0.6[/tex]

[tex]P(R') = 0.4[/tex]

The probability that it'll not rain during the next 3 days is:

[tex]P(R') * P(R') * P(R')[/tex]

[tex]P(R') * P(R') * P(R') =0.4 * 0.4 * 0.4[/tex]

[tex]P(R') * P(R') * P(R') = 0.064[/tex]

Answer:

Pretty sure it is c

Step-by-step explanation:

Answer:

C.

Step-by-step explanation:

She will be painting the outsides of the table, so we need to find the surface area of the table.

There is the flat part of the table, which is a rectangular prism. There are also four legs, which are rectangular prism.

So, she will paint C. the surface area of 6 rectangular prisms.

Hope this helps!

Answer:

Area of the common pasture = 12 acres

Step-by-step explanation:

Let the dimensions of the farm owned by Jay are 'a' units and 'b' units.

Area of the farm = ab = 2 acres

Similarly, areas of the farm owned by Peter with dimensions 'a' unit and 'c' unit = ac = 4 acres

And area of the farm owned by Sam with dimensions 'b' and 'd' units = bd = 6 acres

Now, [tex]\frac{ab}{ac}=\frac{2}{4}[/tex]

[tex]\frac{b}{c}=\frac{1}{2}[/tex] ---------(1)

[tex]\frac{ab}{bd}=\frac{2}{6}[/tex]

[tex]\frac{a}{d}=\frac{1}{3}[/tex] ---------(2)

[tex]\frac{b}{c}\times \frac{a}{d}=\frac{1}{2}\times \frac{1}{3}[/tex]

[tex]\frac{ab}{cd}=\frac{1}{6}[/tex]

cd = 6(ab)

cd = 6 × 2 [Since ab = 2 acres]

    = 12 acres

Therefore, area of the common pasture will be 12 acres.

Answer:

Step-by-step explanation:

Hello!

Regarding the reasons that traffic accidents occur:

28% are caused by distracted drivers using cell phones or texting

11% of the drivers' user their phones at any time

The probability of a driver having an accident is 5.26%

a)

DC = event that a randomly selected driver is using a cell phone.

P(DC)= 0.11

b)

TA = event that a randomly selected driver has a traffic accident.

P(TA)= 0.0526

c) and f)

If both events are related, i.e. dependent, then you would expect that the occurrence of one of these events will affect the probability of the other one. If they are not related, i.e. independent events, then their probabilities will not be affected by the occurrence of one or another:

If both events are independent P(TA|DC)= P(TA)

If they are dependent, then:

P(TA|DC)≠ P(TA)

P(TA|DC)= 0.28

P(TA)= 0.0526

As you can see the probability of the driver having an accident given that he was using the cell phone is different from the probability of the driver having an accident. This means that both events are related.

d) and e)

You have to calculate the probability that "the driver was distracted with the phone given that he had an accident", symbolically P(DC|TA)

P(DC|TA) = [tex]\frac{P(DCnTA)}{P(TA)}[/tex]

[tex]P(TA|DC)= \frac{P(TAnDC}{P(DC)}[/tex] ⇒ P(DC∩TA)= P(TA|DC)*P(DC)= 0.28 *  0.11= 0.0308

P(DC|TA) = [tex]\frac{0.0308}{0.0526}= 0.585= 0.59[/tex]

I hope this helps!

Answer:

The variance of the data is 15.

σ² = 15

Step-by-step explanation:

The mean is given as

X = 8

х        |    (x - X)    |    (x - X) ²

12       |        4         |    16

9        |        1         |     1    

11        |        3         |    9

5       |        -3        |    9

3       |        -5        |    25

The variance is given by

[tex]\sigma^2 = \frac{1}{n-1} \sum (x - X)^2[/tex]

[tex]\sigma^2 = \frac{1}{5 - 1} (16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} ( 16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} (60) \\\\\sigma^2 = 15[/tex]

Therefore, the variance of the data is 15.

Answer:

$10 + $10 + $1 + 25¢ + 5¢

or

$20 + $1 + 25¢ + 5¢

Step-by-step explanation:

Each one must pay $21.30

Answer:

Length of Shortest Side = 12 inches

Step-by-step explanation:

Length of Shortest Side = L = 3x

Length of Longest Side = W = 5x-4

Condition:

2L+2W = Perimeter

2(3x)+2(5x-4) = 56

6x+10x-8 = 56

16x-8 = 56

Adding 8 to both sides

16x = 56+8

16x = 64

Dividing both sides by 14

=> x = 4

Now,

Length of the Shortest Side = L = 3(4) = 12 inches

Length of the Longest Side = W = 5(4)-4 = 16 inches

Answer:

12 inches

Step-by-step explanation:

The length is the longest side.

The width is the shortest side.

Length : [tex]l=5x-4[/tex]

Width : [tex]w=3x[/tex]

Apply formula for the perimeter of a rectangle.

[tex]P=2l+2w[/tex]

[tex]P=perimeter\\l=length\\w=width[/tex]

Plug in the values.

[tex]56=2(5x-4)+2(3x)[/tex]

[tex]56=10x-8+6x[/tex]

[tex]56=16x-8[/tex]

[tex]64=16x[/tex]

[tex]4=x[/tex]

The shortest side is the width.

[tex]w=3x[/tex]

Plug in the value for x.

[tex]w=3(4)[/tex]

[tex]w=12[/tex]

Answer:

  ∞

Step-by-step explanation:

c can have any value you like.

There is no maximum. We say it can approach infinity.

__

Additional comment

There may be some maximum imposed by constraints not shown here. Since we don't know what those constraints are, we cannot tell you what the maximum is.

Complete Question

According to a study done by the Gallup organization, the proportion of Americans who are satisfied with the way things are going in their lives is 0.82. Suppose a random sample of 100 Americans is asked "Are you satisfied with the way things are going in your life?"

What is the probability the sample proportion who are satisfied with the way things are going in their life is greater than 0.85

Answer:

The probability is  [tex]P(X > 0.85 ) = 0.21745[/tex]

Step-by-step explanation:

From the question we are told that

   The population proportion is [tex]p = 0.82[/tex]

   The value considered is  x  =  0.85

     The  sample size is  n = 100

The standard deviation for this population proportion is evaluated as

        [tex]\sigma = \sqrt{\frac{p(1-p)}{n} }[/tex]

substituting values

       [tex]\sigma = \sqrt{\frac{0.82(1-0.82)}{100} }[/tex]

      [tex]\sigma = 0.03842[/tex]

Generally the probability that probability the sample proportion who are satisfied with the way things are going in their life is greater than x is mathematically represented as

       [tex]P(X > x ) = P( \frac{X - p }{ \sigma } > \frac{x - p }{ \sigma } )[/tex]

Where  [tex]\frac{X - p }{ \sigma }[/tex] is  equal to Z (the  standardized value of X ) so  

         [tex]P(X > x ) = P( Z> \frac{x - p }{ \sigma } )[/tex]

substituting values

        [tex]P(X > 0.85 ) = P( Z> \frac{ 0.85 - 0.82 }{ 0.03842 } )[/tex]

        [tex]P(X > 0.85 ) = P( Z> 0.78084)[/tex]

from the standardized normal distribution table [tex]P( Z> 0.78084)[/tex] is  0.21745

So  

      [tex]P(X > 0.85 ) = 0.21745[/tex]

Answer:

The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.

Step-by-step explanation:

Vectorially speaking, let assume that ship A is located at the origin and the relative distance of ship B with regard to ship A at noon is:

[tex]\vec r_{B/A} = \vec r_{B} - \vec r_{A}[/tex]

Where [tex]\vec r_{A}[/tex] and [tex]\vec r_{B}[/tex] are the distances of ships A and B with respect to origin.

By supposing that both ships are travelling at constant speed. The equations of absolute position are described below:

[tex]\vec r_{A} = \left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i[/tex]

[tex]\vec r_{B} = \left(170\,km\right)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j[/tex]

Then,

[tex]\vec r_{B/A} = (170\,km)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j-\left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i[/tex]

[tex]\vec r_{B/A} = \left[170\,km-\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j[/tex]

The rate of change of the distance between the ship is constructed by deriving the previous expression:

[tex]\vec v_{B/A} = -\left(40\,\frac{km}{h} \right)\cdot i + \left(15\,\frac{km}{h} \right)\cdot j[/tex]

Its magnitude is determined by means of the Pythagorean Theorem:

[tex]\|\vec v_{B/A}\| = \sqrt{\left(-40\,\frac{km}{h} \right)^{2}+\left(15\,\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec r_{B/A}\| \approx 42.720\,\frac{km}{h}[/tex]

The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.

Answer:

(a) A 95​% confidence interval estimate of the percentage of orders that are not accurate is [0.125, 0.201].

(b) We can conclude that both restaurants can have the same inaccuracy rate due to the overlap of interval areas.

Step-by-step explanation:

We are given that in a study of the accuracy of fast food​ drive-through orders, Restaurant A had 302 accurate orders and 59 orders that were not accurate.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                          P.Q.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of orders that were not accurate = [tex]\frac{59}{361}[/tex] = 0.163

          n = sample of total orders = 302 + 59 = 361

          p = population proportion of orders that are not accurate

Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                    of significance are -1.96 & 1.96}  

P(-1.96 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.96) = 0.95

P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95

P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95

95% confidence interval for p = [ [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]

  = [ [tex]0.163 -1.96 \times {\sqrt{\frac{0.163(1-0.163)}{361} } }[/tex] , [tex]0.163 +1.96 \times {\sqrt{\frac{0.163(1-0.163)}{361} } }[/tex] ]

  = [0.125, 0.201]

(a) Therefore, a 95​% confidence interval estimate of the percentage of orders that are not accurate is [0.125, 0.201].

(b) We are given that the 95​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B is [0.143 < p < 0.219].

Here we can observe that there is a common area of inaccurate order of 0.058 or 5.85% for both the restaurants.

So, we can conclude that both restaurants can have the same inaccuracy rate due to the overlap of interval areas.

Answer:

75%

Step-by-step explanation:

First we can solve the area of the rectangle originally the answer is:

4 × 2 = 8

Then we decrease both measurements by 50% to get the dimensions 1 and 2. The new area will be 1 × 2 which is 2.

2 is 25% of 8 which means that the area of the rectangle has been reduced by 75%.