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3 years ago

Figure f is a scaled copy of Figure e.

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

4 0

The correct equation of the scaled figure is given by 6/3 = 4/a

<h3>Scaling</h3>

Scaling is the process of either increasing or decreasing the size of a figure by a factor.

From the diagram attached, Figure f is a scaled copy of Figure e. Figure e is decreased to produce figure f, hence:

AB/XY = CD / WZ
6/4 = 3 / a
6/3 = 4/a

The correct equation of the scaled figure is given by 6/3 = 4/a

Find out more on scaling at: brainly.com/question/10253650

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Apply the distributive property to create an equivalent expression. 2(3-8y)

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Using the distributive property, an equivalent expression would be 6-16y

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The sum of four times a number and 18 is the same as the difference of fourteen and twice a number

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$4x+18=14-2x \\ 4x=-4-2x \\ 6x=-4 \\ x= \frac{2}{3}$

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3 years ago

132=-6+3(1-5p) solve the equation

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Answer:

p=-9

Step-by-step explanation:

132=-6+3 (1-5p)

-6+3 (1-5p)=132

-6+3 (1-5p)+6=132+6

3(1-5p)=138

3(1-5p)/3=138/3

1-5p=46

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-5p=45

p=-9

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3 years ago

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Liliana read two books in three months. What was her rate of reading, in books per month? Give your answer as a whole number or

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Answer:

two thirds

Step-by-step explanation:

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3 years ago

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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

3 years ago