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3 years ago

Round to the nearest tenths Plzz show step by step

Mathematics

1 answer:

zhuklara [117]3 years ago

8 0

So first, separate the triangle from the rectangle. Then calculate the area of the rectangle, length X width. 24 times 8 = 192. Now, for the triangle, we must first find its base. You see the two 6 yd's ??? We add that and we subtract that from 24. We get 12. Since we have to multiply the base times height and divide it by 2 (this is the formula for a triangle), we must find the triangle's height. You can see the red dash going through the shape and it says 16. To the right side of that, we see 8, so we instantly know that the half of that red dash is 8yd. So the triangle's height is 8yd. Multiply 12 by 8 =96. Divide it by two because a triangle is the half of a square, then we get 48yd. Add 48 by 192 and we get 240. The area of this shape is 240 yd.

That was long

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Answer:

The second choice is correct. (f(x) = 8.50x+7.50)

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2 years ago

What is the center of the circle? Please show steps!

Vika [28.1K]

Answer:

Step-by-step explanation:

The equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

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3 years ago

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gtnhenbr [62]

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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)

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