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Dafna1 [17]
2 years ago
5

How do I simplify the radical expression. √8-4√2

Mathematics
1 answer:
WITCHER [35]2 years ago
3 0
√8 = √(4x2)  = 2√2

4√2 =4√2

2√2-4√2 = -2√2


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Please need help on this one
kakasveta [241]

Answer:

25 degrees

Step-by-step explanation:

Use a trig. ratio

sin^{-1}=3/7

Plug it into a calculator

sin^{-1}=25.3

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3 years ago
Find the median for 8, 11, 6, 14, x, 13, and 11
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Median = 11. Sorry if wrong


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4 0
3 years ago
The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)

Calculation the value from standard normal z table, we have,  

P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

Calculating the value from the standard normal table we have,

1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

Calculation the value from standard normal z table, we have,  

P( z \leq -1.036) = 0.15

\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0
2 years ago
You have been asked to design a can shaped like right circular cylinder that can hold a volume of 432π-cm3. What dimensions of t
rosijanka [135]

Answer:

Height = 12cm

Radius = 6cm

Step-by-step explanation:

Given

Represent volume with v, height with h and radius with r

V = 432\pi

Required

Determine the values of h and r that uses the least amount of material

Volume is calculated as:

V = \pi r^2h\\

Substitute 432π for V

432\pi = \pi r^2h

Divide through by π

432 = r^2h

Make h the subject:

h = \frac{432}{r^2}

Surface Area (A) of a cylinder is calculated as thus:

A=2\pi rh+2\pi r^2

Substitute \frac{432}{r^2} for h in A=2\pi rh+2\pi r^2

A=2\pi r(\frac{432}{r^2})+2\pi r^2

A=2\pi (\frac{432}{r})+2\pi r^2

Factorize:

A=2\pi (\frac{432}{r} + r^2)

To minimize, we have to differentiate both sides and set A' = 0

A'=2\pi (-\frac{432}{r^2} + 2r)

Set A' = 0

0=2\pi (-\frac{432}{r^2} + 2r)

Divide through by 2\pi

0= -\frac{432}{r^2} + 2r

\frac{432}{r^2} = 2r

Cross Multiply

2r * r^2 = 432

2r^3 = 432

Divide through by 2

r^3 = 216

Take cube roots of both sides

r = \sqrt[3]{216}

r = 6

Recall that:

h = \frac{432}{r^2}

h = \frac{432}{6^2}

h = \frac{432}{36}

h = 12

Hence, the dimension that requires the least amount of material is when

Height = 12cm

Radius = 6cm

3 0
3 years ago
7th grade work really need help on this!
seropon [69]

I really don’t know

I really dont know

4 0
2 years ago
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