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Bond [772]
3 years ago
8

find two numbers such that the sum of twice the first and thrice the second is 92 ,and four times the first exceeds seven times

the second by 2​
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

First number = 25

Second number  = 14

Step-by-step explanation:

Let the numbers be x & y

The sum of twice the first and thrice the second is 92

2x + 3y = 92   -------------------(I)

Four times the first exceeds seven times the second by 2​

4x - 7y = 2  --------------------(II)

Multiply  equation (I) by -2 and then add the equations

(I)*-2          -4x - 6y = -184

(II)              <u> 4x - 7y = 2  </u>    { Now add, thus x will be eliminated}

                      - 13y = -182 {Divide both sides by -13}

y = -182/-13

y = 14

Plugin the value of y in equation (I)

2x + 3*14 = 92

2x + 42 = 92

       2x = 92 - 42

       2x = 50

        x  = 50/2

x = 25

First number = 25

Second number  = 14

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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
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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.

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