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Rina8888 [55]
3 years ago
5

Question 13 pts

Mathematics
1 answer:
oee [108]3 years ago
7 0

Answer:

1. 68%

2. 50%

3. 15/100

Step-by-step explanation:

Here, we want to use the empirical rule

1. % waiting between 15 and 25 minutes

From what we have in the question;

15 is 1 SD below the mean

25 is 1 SD above the mean

So practically, we want to calculate the percentage between;

1 SD below and above the mean

According to the empirical rule;

1 SD above the mean we have 34%

1 SD below, we have 34%

So between 1 SD below and above, we have

34 + 34 = 68%

2. Percentage above the mean

Mathematically, the percentage above the mean according to the empirical rule for the normal distribution is 50%

3. Probability that someone waits less than 5 minutes

Less than 5 minutes is 3 SD below the mean

That is 0.15% according to the empirical rule and the probability is 15/100

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1/5x-2/3y=30<br> What is the value of the x in the equation when y=15
svetlana [45]

Answer:

\frac{1}{5} x -  \frac{2}{3} y = 30 \\  \frac{1}{5} x -  \frac{2}{3} 15 = 30 \\  \frac{1}{5} x -  \frac{30}{3}  = 30 \\  \frac{1}{5} x  -  10 = 30 \\  \frac{1}{5} x = 30 + 10 \\  \frac{1}{5} x = 40 \\ x = 40 \div  \frac{1}{5}  \\ x = 40 \times  \frac{5}{1}  \\ x = 40 \times 5 \\ x = 200

I hope I helped you^_^

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What is the distance between point C and point D?
e-lub [12.9K]

Answer:

B. 13 units

Step-by-step explanation:

let

(x1 , y1) = (-4 , -6)

(x2 , y2) = (1 , 6)

distance between CD = root (x2 - x1)^3 + (y2 - y1)

= root {1 - ( -4)}^2 + {6 - ( -6)}^2

= root {1 +4}^2 + {6+6}^2

= root 5^2 + 12^2

= root 25 + 144

= root 169

= 13 units ans

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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
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