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4 years ago

Pls help me solve this!

Mathematics

2 answers:

Anika [276]4 years ago

8 0

Answer:

The answer to your question is y = 2/3 x + 2

Step-by-step explanation:

Data

Equation = ?

slope = 2/3

P (-3, -1)

Process

1.- Identify x1 and y1

x1 = -3

y1 = -1

2.- Substitute the values in the equation of the point-slope form.

y - y1 = m (x - x1)

-Substitution

y - (-1) = 2/3(x - (-3))

-Simplification

y + 1 = 2/3x + 3

- Solve for y

y = 2/3 x + 3 - 1

-Simplify

y = 2/3 x + 2

Svet_ta [14]4 years ago

5 0

Answer:

Step-by-step explanation:

The point slope form is expressed as

y - y1 = m(x - x1)

Where

m represents the slope of the line.

The slope of the line is 2/3 and it passes through (- 3, - 1). To determine the equation, we would substitute x1 = - 3, y1 = - 1 and m= 2/3 into the point slope form equation. It becomes

y - - 1 = 2/3(x - - 3)

y + 1 = 2/3(x + 3)

The slope intercept form is expressed as

y = mx + c

Where c represents the y intercept

Therefore,

y + 1 = 2x/3 + 2

y = 2x/3 + 2 - 1

y = 2x/3 + 1

You might be interested in

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)

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