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lys-0071 [83]
3 years ago
7

Assume that the population proportion is 0.56. Compute the standard error of the proportion, σp, for sample sizes of 100, 200, 5

00, and 1,000. (Round your answers to four decimal places.)
Mathematics
1 answer:
Aliun [14]3 years ago
4 0

Answer:

Standard errors are 0.049, 0.035, 0.022, and 0.016.

Step-by-step explanation:

The given value of population proportion (P) = 0.56

Given sample sizes (n ) 100, 200, 500, and 1000.

Now standard error is required to calculate.

Use the below formula to find standard error.

When sample size is n = 100

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{100}} =0.049

When sample size is n = 200

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{200}} = 0.035

When sample size is n = 500

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{500}} =0.022

When sample size is n = 1000

\sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.56(1-0.56)}{1000}} = 0.016

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Evaluate the function.<br> f(x) = -x² + 6<br> Find f(-3)
madam [21]

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f(- 3) = - 3

Step-by-step explanation:

To evaluate f(- 3) substitute x = - 3 into f(x) , that is

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3 years ago
Read 2 more answers
QUESTION 10
adelina 88 [10]

Answer:

Step-by-step explanation:

A = L * W

A = 22

L = 2x + 1

W = 3x + 1

now we sub

22 = (2x + 1)(3x + 1)

22 = 6x^2 + 2x + 3x + 1

22 = 6x^2 + 5x + 1

6x^2 + 5x + 1 - 22 = 0

6x^2 + 5x - 21 = 0 <======part A

(3x + 7)(2x - 3) = 0

3x + 7 = 0          2x - 3 = 0

3x = -7               2x = 3

x = -7/3              x = 3/2

I believe that x = -7/3 is an extraneous solution because when u plug it into ur length and width u get a negative number....and I dont think the sides of ur rectangle have negative values, however, it does work in ur equation and it equals 22 when multiplied...so I am not 100% sure

length = 2x + 1.......2(3/2) + 1........3 + 1 = 4

width = 3x + 1......3(3/2) + 1......9/2 + 1......9/2 + 2/2 = 11/2 (or 5.5) meters <===

6 0
3 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
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