If 14% of a number is 18, find 7% of that number.

Following are the code to this question:

   public double distance(Point next) //defining distance method that accepts Constructor

   {

   int x1,x2,y1,y2;//defining integer variables

   double d;//defining double variable dis

   x1=this.x; //use x1 variable that use this keyword to store x variable value

   y1=this.y;//use y1 variable that use this keyword to store y variable value

   x2=next.x;//use x2 variable that use this keyword to store x variable value

   y2=next.y;//use y1 variable that use this keyword to store y variable value

   d=Math.sqrt((x1-0)*(x1-0) + (y1-0)*(y1-0));//use d variable that calculates distance between p1 to origin and store its value

   d=Math.sqrt((x2-0)*(x2-0) + (y2-0)*(y2-0));//use d variable that calculates distance between p2 to origin and store its value

   d=Math.sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1));//calculating the distance from p1 to p2 and store its value

   return d;//return dis value

   }

Explanation:

In the above-given code, the double type method "distance" is defined, that accepts a constructor in its parameter, and inside the method, four integer variable "x1,x2,y1, and y2" and one double variable "d" is declared.  

After holding the value of x and y into the declared integer variable the "d" variable is used, that uses the maths square method with an integer variable, which holds p1 to origin value, p2 to origin value, and the distance from p1 to p2, and use the return keyword for return its value.

9

Step-by-step explanation:

14%= 18

1%=

7%= x 7

    =

    = 9

Following are the code to this question:

   public double distance(Point next) //defining distance method that accepts Constructor

   {

   int x1,x2,y1,y2;//defining integer variables

   double d;//defining double variable dis

   x1=this.x; //use x1 variable that use this keyword to store x variable value

   y1=this.y;//use y1 variable that use this keyword to store y variable value

   x2=next.x;//use x2 variable that use this keyword to store x variable value

   y2=next.y;//use y1 variable that use this keyword to store y variable value

   d=Math.sqrt((x1-0)*(x1-0) + (y1-0)*(y1-0));//use d variable that calculates distance between p1 to origin and store its value

   d=Math.sqrt((x2-0)*(x2-0) + (y2-0)*(y2-0));//use d variable that calculates distance between p2 to origin and store its value

   d=Math.sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1));//calculating the distance from p1 to p2 and store its value

   return d;//return dis value

   }

Explanation:

In the above-given code, the double type method "distance" is defined, that accepts a constructor in its parameter, and inside the method, four integer variable "x1,x2,y1, and y2" and one double variable "d" is declared.  

After holding the value of x and y into the declared integer variable the "d" variable is used, that uses the maths square method with an integer variable, which holds p1 to origin value, p2 to origin value, and the distance from p1 to p2, and use the return keyword for return its value.

Step-by-step explanation:

step 1

Find the volume of one container (cylinder) is equal to

we have

-----> the radius is half the diameter

substitute

step 2

By proportion

Find the the minimum number of identical containers that Sue would need to make of ice

Round to the nearest whole number

Step-by-step explanation:

step 1

Find the volume of one container (cylinder) is equal to

we have

-----> the radius is half the diameter

substitute

step 2

By proportion

Find the the minimum number of identical containers that Sue would need to make of ice

Round to the nearest whole number

Step-by-step explanation:

H0: u = 150

H1: u ≠ 150

Total sum = 15933.24

N = 100

Mean = 15933.24/100

= 159.3324

σ² = 25954.03 - (159.3324)²/100

σ² = 556.213

σ = √556.213

σ = 23.8162

Testing hypothesis

t = (bar x - u)/ σ/√n

= 159.3324-150/23.8162/√100

= 3.91

We will have a p value of 0.02

0.0002 < 0.01

We reject null hypothesis at 1% level of significance

C. Mean = 159.3324

Se= 2.3936

Df = 100-1 = 99

Critical value at 0.01 = +-2.626

T = x-u/s.e

= -2.626 =( x -150)/2.3936

When we cross multiply and solve this

X = 143.714 for the lower tail

2.626 = (x-159)/2.3936

= 156.286 for upper tail.

We therefore reject H0 at

Bar X <= 143.71

Bar X >= 156.286

At 10%

Critical t = 1.660

-1.660 = (x - 150)/2.3936

Solving this ,

X =146.02 at the lower tail

1.660 = (x-150)/2.3936

X = 153.97

We reject H0 at

X<= 146.03

X>=153.97

Step-by-step explanation:

H0: u = 150

H1: u ≠ 150

Total sum = 15933.24

N = 100

Mean = 15933.24/100

= 159.3324

σ² = 25954.03 - (159.3324)²/100

σ² = 556.213

σ = √556.213

σ = 23.8162

Testing hypothesis

t = (bar x - u)/ σ/√n

= 159.3324-150/23.8162/√100

= 3.91

We will have a p value of 0.02

0.0002 < 0.01

We reject null hypothesis at 1% level of significance

C. Mean = 159.3324

Se= 2.3936

Df = 100-1 = 99

Critical value at 0.01 = +-2.626

T = x-u/s.e

= -2.626 =( x -150)/2.3936

When we cross multiply and solve this

X = 143.714 for the lower tail

2.626 = (x-159)/2.3936

= 156.286 for upper tail.

We therefore reject H0 at

Bar X <= 143.71

Bar X >= 156.286

At 10%

Critical t = 1.660

-1.660 = (x - 150)/2.3936

Solving this ,

X =146.02 at the lower tail

1.660 = (x-150)/2.3936

X = 153.97

We reject H0 at

X<= 146.03

X>=153.97

u what kind of math are u doing

Step-by-step explanation:

1) a. CI = 0.426 < p < 0.574

2) c. CI = 2.74% < p < 6.524%

3) a. CI = 20 < μ < 22

4) c. CI = 81.11 < μ < 85.09

5) b. CI = 76.21 < μ < 89.79

6) d. $453.59 < μ < $874.69

7) d. CI = $559 < σ < $1953.3

8) c. CI = $3.96 < σ < $6.72

9) c. 6.22 ft < σ < 12.59 ft

10) d. 0.19 oz < σ < 0.72 oz

Step-by-step explanation:

The Confidence Interval, CI are given as follows

For the population proportion

1)   = 55/110 = 0.5

z at 88% = ±1.56

a. CI = 0.426 < p < 0.574

2)   = 31/669 = 0.5

z at 98% = ±2.326

c. CI = 2.74% < p < 6.524%

3) Here we have unknown population standard deviation, so we find the t interval

= 21

n = 130

s = 3.0

Confidence level = 98%

= ±2.356

CI = 20.380 < μ < 21.62

Rounding up gives;

a. CI = 20 < μ < 22

4) Similarly here we have;

= ±1.699 and

c. CI = 81.11 < μ < 85.09

5) Here

= 83

s = 13.5

n = 30

Confidence level = 99%

= ±2.76 and

b. CI = 76.21 < μ < 89.79

6) In the question, we have

= $664.14

s = $297.29

n = 14

Confidence level = 98%

= ±2.65 and

CI = $453.56 < μ < $874.72

d. $453.59 < μ < $874.69

7) The confidence interval for a population standard deviation is given by the following relation;

Here

n = 9

x = $3959

s = $886

Therefore we have

d. CI = $559 < σ < $1953.3

8) Here we have;

n = 41

x = $108

s = $5

Therefore;

c. CI = $3.96 < σ < $6.72

9) Here we have;

n = 29

x = 65 ft

s = 8.5 ft

Therefore we have

CI = 6.3 ft < σ < 12.73 ft

c. 6.22 ft < σ < 12.59 ft

10) Here we have

n = 8

s = 0.301188123

d. 0.19 oz < σ < 0.72 oz.

1) a. CI = 0.426 < p < 0.574

2) c. CI = 2.74% < p < 6.524%

3) a. CI = 20 < μ < 22

4) c. CI = 81.11 < μ < 85.09

5) b. CI = 76.21 < μ < 89.79

6) d. $453.59 < μ < $874.69

7) d. CI = $559 < σ < $1953.3

8) c. CI = $3.96 < σ < $6.72

9) c. 6.22 ft < σ < 12.59 ft

10) d. 0.19 oz < σ < 0.72 oz

Step-by-step explanation:

The Confidence Interval, CI are given as follows

For the population proportion

1)   = 55/110 = 0.5

z at 88% = ±1.56

a. CI = 0.426 < p < 0.574

2)   = 31/669 = 0.5

z at 98% = ±2.326

c. CI = 2.74% < p < 6.524%

3) Here we have unknown population standard deviation, so we find the t interval

= 21

n = 130

s = 3.0

Confidence level = 98%

= ±2.356

CI = 20.380 < μ < 21.62

Rounding up gives;

a. CI = 20 < μ < 22

4) Similarly here we have;

= ±1.699 and

c. CI = 81.11 < μ < 85.09

5) Here

= 83

s = 13.5

n = 30

Confidence level = 99%

= ±2.76 and

b. CI = 76.21 < μ < 89.79

6) In the question, we have

= $664.14

s = $297.29

n = 14

Confidence level = 98%

= ±2.65 and

CI = $453.56 < μ < $874.72

d. $453.59 < μ < $874.69

7) The confidence interval for a population standard deviation is given by the following relation;

Here

n = 9

x = $3959

s = $886

Therefore we have

d. CI = $559 < σ < $1953.3

8) Here we have;

n = 41

x = $108

s = $5

Therefore;

c. CI = $3.96 < σ < $6.72

9) Here we have;

n = 29

x = 65 ft

s = 8.5 ft

Therefore we have

CI = 6.3 ft < σ < 12.73 ft

c. 6.22 ft < σ < 12.59 ft

10) Here we have

n = 8

s = 0.301188123

d. 0.19 oz < σ < 0.72 oz.