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Vika [28.1K]
3 years ago
9

What is 5/9*2 4/ equal

Mathematics
1 answer:
Kobotan [32]3 years ago
6 0
The answer is 1.11111
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Use Random number generator and simulate 1000 long columns, for each of the three cases. Example: for the Car type 1, use Number
sertanlavr [38]

Answer:

Step-by-step explanation:

The question is incomplete since they do not give information about the Car type 3.

We will do it in a generic way, we will say that the Car type 3 has a mean of M and a standard deviation SD.

We would be:

P (CT3 <550) = P [z <(550 - X) / SD]

Now if we give it values, for example that X = 600 and SD = 120

It would remain:

P (CT3 <550) = P [z <(550 - 600) / 120]

P (CT3 <550) = P [z <-0.42]

We look for this value in the normal distribution table (attached) and it shows us that the probability is approximately 0.3372, that is, 33.72%

What you need to do is replace the X and SD values of theCar type 3 in the equation above how I just did and you will get the result.

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2 years ago
The answer must be a mixed number a proper fraction or a integer
mina [271]

Answer:

i dont understand sir sorry

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2 years ago
Z-2w=2 please help me solve , studying for praxis
RUDIKE [14]

Answer:

Z(2w-2)=24

Simple and best practice solution for Z(2w-2)=24 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.

5 0
2 years ago
If it takes you 20 minutes to hike a distance of 1 mile, how fast are you walking in miles per hour
Leni [432]

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1.60934 per hour

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4 0
3 years ago
Read 2 more answers
Which of the following are square roots of —8 + 8i/3? Check all that apply.
8090 [49]

Answer:

Options (2) and (3)

Step-by-step explanation:

Let, \sqrt{-8+8i\sqrt{3}}=(a+bi)

(\sqrt{-8+8i\sqrt{3}})^2=(a+bi)^2

-8 + 8i√3 = a² + b²i² + 2abi

-8 + 8i√3 = a² - b² + 2abi

By comparing both the sides of the equation,

a² - b² = -8 -------(1)

2ab = 8√3

ab = 4√3 ----------(2)

a = \frac{4\sqrt{3}}{b}

By substituting the value of a in equation (1),

(\frac{4\sqrt{3}}{b})^2-b^2=-8

\frac{48}{b^2}-b^2=-8

48 - b⁴ = -8b²

b⁴ - 8b² - 48 = 0

b⁴ - 12b² + 4b² - 48 = 0

b²(b² - 12) + 4(b² - 12) = 0

(b² + 4)(b² - 12) = 0

b² + 4 = 0 ⇒ b = ±√-4

                     b = ± 2i

b² - 12 = 0 ⇒ b = ±2√3

Since, a = \frac{4\sqrt{3}}{b}

For b = ±2i,

a = \frac{4\sqrt{3}}{\pm2i}

  = \pm\frac{2i\sqrt{3}}{(-1)}

  = \mp 2i\sqrt{3}

But a is real therefore, a ≠ ±2i√3.

For b = ±2√3

a = \frac{4\sqrt{3}}{\pm 2\sqrt{3}}

a = ±2

Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)

Options (2) and (3) are the correct options.

6 0
3 years ago
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