look at the picture
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The number of the sample that is necessary is 171. The number of employees that would be selected if E is 30 = 476

The standard deviation = 398

error E = 50

Alpha is = 1-0.90 = 0.10

a. We are to find the value of the size of the sample

Z∝/2 = Z0.05

= 1.645

sample size,n >= [Za/2 *(sigma/E)]^2

n >= [1.645 * (398/50)] ^2

n >= 171.46

n = 171

b.

error,E = 30

n >= [1.645 * (398/30)] ^2

n >= 476.27

n = 476

A survey is planned to determine the mean annual family medical expenses of employees of a large company. The management of the company wishes to be 90% confident that the sample mean is correct to within ±$50 of the population mean annual family medical expenses. A previous study indicates that the standard deviation is approximately $398. a. How large a sample is necessary?

b. If management wants to be correct to within ±$30, how many employees need to be selected?

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Answer:

Step-by-step explanation:

First, lets open up the parentheses.

3x - 6x + 4 = 8 - 2x + 3

Combine like terms.

-3x + 4 = 11 - 2x

Add 2x to both sides

-x + 4 = 11

Subtract 4 from both sides

-x + 4 -4 = 11 -4

Simplify

-x = 7

Divide both sides by -1

x = -7

Hope this helps! (:

Answer:

x = -7

Step-by-step explanation:

3x – 2(3x – 2) = 2(4 – x) + 3

The first step is to distribute when solving for x

3x – 6x +4 = 8 – 2x + 3

Then combine like terms

-3x+4=-2x+11

Add 3x to each side

-3x+3x+4=-2x+3x+11

4 = x+11

Subtract 11 from each side

4-11 = x+11-11

-7 = x

Answer:

5.5 in

Step-by-step explanation:

First we will calculate the total lawn grown after 8 weeks.

Grow Rate = [tex]2\frac{1}{8} in/2 weeks[/tex]

Total lawn grown after 8 weeks = [tex]2\frac{1}{8} *4\\=8.5 in[/tex]

Next we will calculate the total lawn mown after 8 weeks.

Mowing Rate = [tex]1\frac{3}{4} in/2weeks[/tex]

Total lawn mown after 8 weeks = [tex]1\frac{3}{4} *4\\=7 in[/tex]

Lawn Height after 8 weeks = Initial Lawn Height + Total lawn grown - Total lawn mown

= 4 in + 8.5 in - 7 in

= 5.5 in

Answer: 30

Step-by-step explanation: Formula for Area of a triangle = 1/2 Base * Height. 12*5 = 60; 60/2 = 30.

Answer:

30 cm squared

Step-by-step explanation:

This is a right angle triangle. That means that the base and height will just be the perpendicular sides of the triangle.

Our base in this case is 12 and our height is 5.

Area of a Triangle = 1/2(base)(height)

=1/2(12)(5)

=1/2(60)

=30

So, the area of this triangle is 30 cm squared.

Answer:

False

Step-by-step explanation:

The slope is 0.

Answer:

Hay 600 en el club de magia.

Answer:

Step-by-step explanation:

The distance between the poles is the segment d which connects the top of the shorter pole with the point at 5 m from the top of the longer pole, which is the difference of lengths of poles.

The three points form a right triangle, with hypotenuse of 16 m, and legs 5 m and d.

Use Pythagorean to find the value of d:

Option c) 2x-3y ≥ 3 is correct inequality which has a shaded area below the boundary line.

Attempting all the equation on the graph to get the solutions

Inequality can be define as the relation of equation contains the symbol of ( ≤, ≥, <, >) instead of equal sign in an equation.

Since, while attempting all the option on graph. The one equation that satisfies the condition is 2x-3y ≥ 3 of having shaded area below the boundary line.

Thus, the option c) 2x-3y ≥ 3 is correct implification on graph which shows the shaded area below the boundary line.

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Answer:

2. (3+4) should be (3*4)

3. x is also x^1, so it should be x^(6+3+1)

4. you have to do 3^4 and 2^3 before timing them together

Answer:

14

Step-by-step explanation:

CB/AB = CE/ET

3/7 = 6/ET

ET = 14

Answer is (4,7)

Step-by-step explanation:

Answer:

yes I think I'm new to this but I think that's a yes

In 11.76 months, the value of the population will be 15000.

We have given that,

At the end of January 2003, the population of my town was 11,212.

the population is increased by 322 every month

So at the end of the second-month population will be 11534.

suppose x represents the year and  y represent the population

The population increased by 322 every month

Therefore the value of m =322

[tex](x_1,y_1)=(0,11212)[/tex]

We have m=332

[tex](x_2,y_2)=(x_2,15000)[/tex]

We have to determine the value of the x_2

[tex](y_2-y_1)=m(x_2-x_1)[/tex]

[tex]322\left(x_2-0\right)=15000-11212[/tex]

[tex]322\left(x_2-0\right)=3788[/tex]

[tex]\frac{322\left(x_2-0\right)}{322}=\frac{3788}{322}[/tex]

[tex]x_2=\frac{1894}{161}[/tex]

[tex]x_2=11.76[/tex]

Therefore in 11.76 months, the value of the population will be 15000.

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Answer:

170.833

Step-by-step explanation:

we have to find real value which has to be find.

let take it as x,

in the question they said reduction percentage is equal to $123

then x*72/100=$123

x=$123*100/72

x=12300/72

x=170.8333

Answer:

I believe it's 211.90$.

The first ten terms of the sequence [tex]a_n=1 + (-\frac{4}{9} )^n[/tex] is 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003

An equation is an expression that shows the relationship between two or more variables and numbers.

Given that:

[tex]a_n=1 + (-\frac{4}{9} )^n\\\\When\ n=1:a_1=1 + (-\frac{4}{9} )^1=0.5556\\\\When\ n=2:a_2=1 + (-\frac{4}{9} )^2=1.1975\\\\When\ n=3:a_3=1 + (-\frac{4}{9} )^3=0.9122\\\\When\ n=4:a_4=1 + (-\frac{4}{9} )^4=1.039\\\\When\ n=5:a_5=1 + (-\frac{4}{9} )^5=0.9827\\\\When\ n=6:a_6=1 + (-\frac{4}{9} )^6=1.0077\\\\When\ n=7:a_7=1 + (-\frac{4}{9} )^7=0.9966\\\\When\ n=8:a_8=1 + (-\frac{4}{9} )^8=1.0015\\\\When\ n=9:a_9=1 + (-\frac{4}{9} )^9=0.9993\\\\[/tex]

[tex]When\ n=10:a_{10}=1 + (-\frac{4}{9} )^{10}=1.0003\\[/tex]

The first ten terms of the sequence [tex]a_n=1 + (-\frac{4}{9} )^n[/tex] is 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003

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The volume of the square prism shown is 1568 cubic meters

The formula for calculating the volume of square prism is expressed as;

V = Base area * height/3

Determine the base are

B = 14 * 14

B = 196 square meters

Determine the height

h = √25² - 7²

h = √625 - 49
h = 24m

Substitute

Volume = 196. * 24/3

Volume = 196 * 8

Volume = 1568 cubic meters

Hence the volume of the square prism shown is 1568 cubic meters

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Answer:

960

Step-by-step explanation:

(if i made any typos, sorry it is kind of hard to type on computer over writing on paper.)

A=2AB+(a+b+c)h

AB=s(s﹣a)(s﹣b)(s﹣c)

s=(a+b+c)/2

Solving forA

A=ah+bh+ch+1/2 sqrt(﹣a^4+2(ab)^2+2(ac)^2﹣b4+2(bc)^2﹣c^4)=10·12+24·12+26·12+1

2·﹣104+2·(10·24)2+2·(10·26)2﹣244+2·(24·26)2﹣264=960

[tex]im \: guessing \: its \: 4x {}^{2} - 3x + 1 \\ (correct \: me \: if \: im \: wrong)[/tex]

[tex](a - b) {}^{2} = a {}^{2} - 2ab + {b}^{2} [/tex]

[tex]4x {}^{2} = a {}^{2} \: \: \: \: \: \: \: \: \: \: \: b {}^{2} = 1 \\ a = 2x \: \: \: \: \: \: \: \: \: \: \: \: \: \: b = 1[/tex]

[tex](2x - 1) {}^{2} = 4x {}^{2} -4x + 1[/tex]

[tex]we \: need \: to \: add \: a \: - x \\ \: to \: the \: original \: equation[/tex]

The  journal entries relating to the building for the fifth year is: Debit Depreciation expense   $10,500; Credit To accumulated depreciation        $10,500.

Bailand Company

1. Dec 31  

Debit Depreciation expense         $10,500

Credit Accumulated depreciation                     $10,500

(Being the depreciation expense recorded)

Book value=$148,000-($148,000-$8,000/10×4)]

Book value=$148,000-$56,000

Book value=$92,000

Depreciation=$92,000-$8,000/8

Depreciation=$10,500

2. Dec 31  

Debit Depreciation expense $24,000.00

Credit Accumulated depreciation a/c               $24,000.00

[($92,000-$8,000)×6/21]

(Being the depreciation expense recorded)

3. Dec 31    

Debit Accumulated depreciation $3,200.00

[($8,000×4)/10]

Credit  Retained earnings              $3,200.00

(Being prior year adjustment for depreciation expense)

Dec 31  

Debit  Depreciation expense         $10,000.00

Credit Accumulated depreciation             $10,000.00

[($148,000-$8,000)/10]

(Being the depreciation expense recorded)

Therefore the  journal entries relating to the building for the fifth year is: Debit Depreciation expense   $10,500; Credit To accumulated depreciation   $10,500.

The complete question is:

Bailand Company purchased a building for $148,000 that had an estimated residual value of $8,000 and an estimated service life of 10 years. Bailand purchased the building 4 years ago and has used straight-line depreciation. At the beginning of the fifth year (before it records depreciation expense for the year), the following independent situations occur:

1. Bailand estimates that the asset has 8 years’ life remaining (for a total of 12 years).

2. Bailand changes to the sum-of-the-years’-digits method.

3. Bailand discovers that the estimated residual value has been ignored in the computation of depreciation expense.

Required: For each of the independent situations, prepare all the journal entries relating to the building for the fifth year. Ignore income taxes.

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Answer:

Infinite Number of Solutions

Step-by-step explanation:

Hello!

We can solve for y in the first equation, and plug that value in for y in the second equation.

Now that we know the equation for y, we can plug that in for y into the second equation to solve for x.

Since the equations match, there are an infinite number of solutions.

If we take a close look at the equations, we can also see that they are the same equation, but with flipped signs.

Answer:

Like terms

Step-by-step explanation:

Both terms have the same unit, which is pounds.

Answer:

-960

Step-by-step explanation:

1) Expand [tex](x - 2y)^{10}[/tex] using the binomial theorem.

Binomial Theorem: [tex]\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i[/tex]

1.1) Substitute the values into the theorem.

[tex]\sum _{i=0}^{10}\binom{10}{i}x^{\left(10-i\right)}\left(-2y\right)^i[/tex]

1.2) Expand summation.

[tex]=\frac{10!}{0!\left(10-0\right)!}x^{10}\left(-2y\right)^0+\frac{10!}{1!\left(10-1\right)!}x^9\left(-2y\right)^1+\frac{10!}{2!\left(10-2\right)!}x^8\left(-2y\right)^2+\frac{10!}{3!\left(10-3\right)!}[/tex]

[tex]x^7\left(-2y\right)^3+\frac{10!}{4!\left(10-4\right)!}x^6\left(-2y\right)^4+\frac{10!}{5!\left(10-5\right)!}x^5\left(-2y\right)^5+\frac{10!}{6!\left(10-6\right)!}x^4\left(-2y\right)^6+\frac{10!}{7!\left(10-7\right)!}x^3\left(-2y\right)^7+\frac{10!}{8!\left(10-8\right)!}x^2\left(-2y\right)^8+\frac{10!}{9!\left(10-9\right)!}x^1\left(-2y\right)^9+\frac{10!}{10!\left(10-10\right)!}x^0\left(-2y\right)^{10}[/tex]

1.3) Simplify them.

1.4) You will get:

[tex]=x^{10}-20x^9y+180x^8y^2-960x^7y^3+3360x^6y^4-8064x^5y^5+13440x^4y^6-15360x^3y^7+11520x^2y^8-5120xy^9+1024y^{10}[/tex]

2) We are told to find the coefficient of [tex]x^{7} y^{3}[/tex]. Find it from the simplified expansion. The coefficient is -960.

The graph of the function. f(x)=−|x+2|+1 is illustrated.

The given function is f(x)=−|x+2|+1.

We will have to draw the graph of the function.

When x = 1, f(1) will be -2. When x = 2, f(2,) will be -3

The points are illustrated based on the information given and the appropriate graph is illustrated and attached.

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