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3 years ago

5

You select every tenth student on an alphabetical list of the students in your school. You survey the selected students in their

first-period classes. Is this question bias or not?

1 answer:

GalinKa [24]3 years ago

7 0

Answer:

no it is not biased because it is randomly selected although in alphabetical order.

Step-by-step explanation:

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A high school gym teacher randomly selects a group of two players from three students to demonstrate a basketball drill during c

Tcecarenko [31]

1/3 chance to pick a boy 2/3 chance to pick a girl

5 0

3 years ago

A dozen eggs cost $1.10 in Dover. In Ensley, the eggs cost 10% more than in Dover. Find the price of a dozen eggs in Ensley.

Julli [10]

1.10 multiplied by .10 = .11
1.10 + .11 = 1.21

In Ensley, the price of a dozen eggs costs $1.21.

Hope this helps.

5 0

4 years ago

Please help me giving out 5 points each

Kay [80]

the answer is B...i swear its B

8 0

3 years ago

Read 2 more answers

The sum of Eric’s and Bob’s weights is 9 times the difference of their weights. The positive difference of their weights is also

Pie

Answer:

<u>150</u>

Step-by-step explanation:

So I will shorten Eric, and Bob's names as E, and B.

So the equations are this

B-E = B+E - 240, E as Eric, and B as Bob, and this E+B=9(E-B)

Move the variables and you get

-2E =- 240

Just divide them both by -2

and E=120

So when we know the value of E we can just plug it into the 1st question

120+B=9B+1080

Moves the Variables and numbers to the other side

8B=1200

1200/8= 150

B=150

There is Bob's weight

6 0

2 years ago

Find all the complex roots. Write the answer in exponential form.

dezoksy [38]

We have to calculate the fourth roots of this complex number:

$z=9+9\sqrt[]{3}i$

We start by writing this number in exponential form:

$\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}$ $\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}$

Then, the exponential form is:

$z=18e^{\frac{\pi}{3}i}$

The formula for the roots of a complex number can be written (in polar form) as:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\cdot\lbrack\cos (\frac{\theta+2\pi k}{n})+i\cdot\sin (\frac{\theta+2\pi k}{n})\rbrack\text{ for }k=0,1,\ldots,n-1$

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.

To simplify the calculations, we start by calculating the fourth root of r:

$r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}$

<em>NOTE: It can not be simplified anymore, so we will leave it like this.</em>

Then, we calculate the arguments of the trigonometric functions:

$\frac{\theta+2\pi k}{n}=\frac{\frac{\pi}{2}+2\pi k}{4}=\frac{\pi}{8}+\frac{\pi}{2}k=\pi(\frac{1}{8}+\frac{k}{2})$

We can now calculate for each value of k:

$\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}$ $\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}$ $\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}$ $\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}$

Answer:

The four roots in exponential form are

z0 = 18^(1/4)*e^(i*π/8)

z1 = 18^(1/4)*e^(i*5π/8)

z2 = 18^(1/4)*e^(i*9π/8)

z3 = 18^(1/4)*e^(i*13π/8)

5 0

1 year ago