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2 years ago

5

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

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

If right, pls mark brainliest i wanna get virtuoso.
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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