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zloy xaker [14]
3 years ago
15

What would be the area of the figure

Mathematics
1 answer:
shusha [124]3 years ago
4 0

Answer:

1.3 + 1.3 = 2.6

2.6 x 2.2 = 5.72

5.72 divided by 2 = 2.86

2.86 x 2 = 5.72

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Directed line segment has endpoints P(– 8, – 4) and Q(4, 12). Determine the point that partitions the directed line segment in a
larisa86 [58]

Answer:

Point (1,8)  

Step-by-step explanation:

We will use segment formula to find the coordinates of point that will partition our line segment PQ in a ratio 3:1.

When a point divides any segment internally in the ratio m:n, the formula is:

[x=\frac{mx_2+nx_1}{m+n},y= \frac{my_2+ny_1}{m+n}]

Let us substitute coordinates of point P and Q as:

x_1=-8,

y_1=-4

x_2=4

y_2=12

m=3

n=1

[x=\frac{(3*4)+(1*-8)}{3+1},y=\frac{(3*12)+(1*-4)}{3+1}]

[x=\frac{12-8}{4},y=\frac{36-4}{4}]

[x=\frac{4}{4},y=\frac{32}{4}]

[x=1,y=8]

Therefore, point (1,8) will partition the directed line segment PQ in a ratio 3:1.

   

7 0
3 years ago
The image Of A(a+2b,2a-b) when reflected in the line y=x is A'(b-a,4b-2a). Show that 3a=2b<br>​
lions [1.4K]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,
Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0
3 years ago
What is 1024 1/5 simplified?
frutty [35]
\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\&#10;a^{-{ n}} \implies \cfrac{1}{a^{ n}}&#10;\qquad \qquad&#10;\cfrac{1}{a^{ n}}\implies a^{-{ n}}&#10;\qquad \qquad &#10;a^{{{  n}}}\implies \cfrac{1}{a^{-{{  n}}}}\\\\&#10;-------------------------------\\\\&#10;1024\frac{1}{5}\qquad &#10;\begin{cases}&#10;1024=2^{10}\\&#10;\frac{1}{5}=5^{-1}&#10;\end{cases}\implies 2^{10}5^{-1}
6 0
3 years ago
Easy points... round 12717. to the nearest hundredth​
Neporo4naja [7]
I just agree with the other comment
3 0
3 years ago
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