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OlgaM077 [116]
3 years ago
9

What is ​11918​ as a mixed number in simplest form? Enter your answer in the box.

Mathematics
2 answers:
sveta [45]3 years ago
8 0

Answer:

the answer is 11 1/2.

Step-by-step explanation:

I used a calculator

bixtya [17]3 years ago
6 0

Answer:

266.0 s ≈ 4.4 min

Step-by-step explanation

1) Data

Wr = 43,576 N

F = 11,918 N

V = 713 m/s

t =?

2) Principles and formulas

Impulse and conservation of momentum

I = F.t = Δp

Δp = mΔv

3) Solution

m = Wr / g = 43,576N / 9.8 m/s^2 = 4,446.5 kg

Δp = mΔv => 4,446.5 kg * 713 m/s = 3,170,354,5 N*s

I = F.t = Δp => t = Δp / F = 3,710,354.5 N*s / 11,918N = 266.0 s

t = 266.0 s ≈ 4.4 min

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If you place a 100-foot ladder against the top of a 96-foot building, how many feet will the bottom of the ladder be from the bo
soldier1979 [14.2K]
X^2+y^2=z^2

x^2+96^2=100^2

x=sqrt(100^2-96^2)

x=28 so the distance from the base of the ladder and the base of the ladder is 28ft.

Hope this helps. Any questions please just ask. Thank you.
4 0
3 years ago
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What addition expression is shown on the<br> number line?<br> 8<br> +<br> ?<br> 15
tekilochka [14]

Answer:

Your answer is seven

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3 years ago
Original price: $50; markdown: 2.2%; retail price:
larisa86 [58]
$50x2.2%
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4 0
3 years ago
The number of texts per day by students in a class is normally distributed with a 
kobusy [5.1K]

Answer:

1, 2, 6

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

4)  For x < 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

For x > 100:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

Option 5 is incorrect

6)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5

Since 1.5 is between 1 and 2, option 6 is correct

5 0
3 years ago
Division of fraction 3/5 divided by 10/11
Tema [17]
3/5 ÷ 10/11
multiply by the inverse

3/5 x 11/10 = 33/50
3 0
3 years ago
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