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marishachu [46]
3 years ago
10

Anton has a deck that is 890 cm by 2891 cm. If he wants to add 66 cm. how large would his deck be?

Mathematics
1 answer:
Aliun [14]3 years ago
4 0

Answer:

The answer is 2,763,796 cm². Or 276.3796m², which is the footprint of a nice size house.

Step-by-step explanation:

890cm \times 2891cm = 2572990 {cm}^{2}

890 + 66 = 956

956cm \times 2891cm = 2763796 {cm}^{2}

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Solve the following equation:
Rama09 [41]

Complete the square.

z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0

\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4

Use de Moivre's theorem to compute the square roots of the right side.

w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)

\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2

Now, taking square roots on both sides, we have

z^2 + \dfrac12 = \pm w^{1/2}

z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2

Use de Moivre's theorem again to take square roots on both sides.

w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)

\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}

w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)

\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}

3 0
2 years ago
The cost of 2 television and 3 DVD player is $1,421 . The cost of 1 DVD player is half the cost of 1 television . What is the co
alexgriva [62]
T=cost of one television D= cost of one DVD player 2t+3d= 1421 D=t(1/2) Replace d by t(1/2) 2t+3(t1/2)= 1421 3.5 t = 1421 T= 691.7 $ Good luck
4 0
4 years ago
F) Determine the domain.
Y_Kistochka [10]

Answer:

1

Step-by-step explanation:

you can tell just divide

4 0
2 years ago
Solve for equation Y<br><br> 15x + 10y = 400
pav-90 [236]

Answer:

15x +10y =400

All divide by 5

3x +2y =80

2y=80 - 3x

y=80-3x/2

6 0
3 years ago
A very large data set (N &gt; 10,000) has a mean value of 1.65 units and a standard deviation of 72.26 units. Determine the rang
Romashka [77]

Answer:

z=-0.674

And if we solve for a we got

a=1.65 -0.674*72.26=-47.05

z=0.674

And if we solve for a we got

a=1.65 +0.674*72.26=50.35

So then the limits where 50% of the data lies are -47.05 and 50.35

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(1.65,72.26)  

Where \mu=1.65 and \sigma=72.26

For this case we want the limits for the 50% of the values.

So on the tails of the distribution we need the other 50% of the data, and on ach tail we need to have 25% since the distribution is symmetric.

Lower tail

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.75   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for a we got

a=1.65 -0.674*72.26=-47.05

So the value of height that separates the bottom 25% of data from the top 75% is -47.05.  

Upper tail

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.674

And if we solve for a we got

a=1.65 +0.674*72.26=50.35

So the value of height that separates the bottom 75% of data from the top 25% is 50.35.  

So then the limits where 50% of the data lies are -47.05 and 50.35

6 0
3 years ago
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