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3 years ago

Multiply: (5x2y – 3xy2)(3xy2 + 5x2y) Please help I need to get this one right!

Mathematics

1 answer:

Aliun [14]3 years ago

6 0

Answer:

(1): "*-5x2y" was replaced by "*(-5x2y)".

(2): "x2" was replaced by "x^2". 1 more similar replacement(s).

STEP

Equation at the end of step 1

(3x • (y2)) • (0 - (5x2 • y))

STEP

Equation at the end of step

3xy2 • -5x2y

STEP

Multiplying exponential expressions :

3.1 x1 multiplied by x2 = x(1 + 2) = x3

Multiplying exponential expressions :

3.2 y2 multiplied by y1 = y(2 + 1) = y3

I believe the answer is ( -3•5x3y3)

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1 year ago

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

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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)

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