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3 years ago

Draw graph of straight lines to find approximate solution to these simultaneous equations . y = 3x + 7 and 2y + x = 2

Mathematics

1 answer:

Aliun [14]3 years ago

3 0

Answer:

(-1.714, 1.857)

Step-by-step explanation:

Here, we want to find the approximate solution to the system of linear equations

We start by plotting the graph of the lines

We have the plot as shown below;

the red line represents the first equation while the blue line represents the second equation

We have the solution as the point at which the two lines meet

The solution to this is;

(-1.714, 1.857)

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3 years ago

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Hannah paid $8.63 at the grocery store for beans and potatoes. She bought 4 pounds of beans at $0.62 per pound and 5 pounds of p

irina [24]

Answer:

$1.23

Step-by-step explanation:

First start off by multiplying $0.62 by 4 to find out the total she paid for all four pounds of beans. This would give you $2.48 spent on beans. Subtract $2.48 from the total amount spent ($8.63). You now have $6.15 left on what she could have spent on 5 pounds of potatoes. Do $6.15 divided by 5 to get an answer of $1.23. Hannah paid $1.23 per pound for the potatoes.

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3 years ago

A software company is releasing one of their products on a CD. The manufacturer charges a $5,500 setup fee and $1.00 per CD. App

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5.50 - 1= 4.50

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1,222 + 5,500 = 6,722
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3 years ago

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Components of a certain type are shipped to a supplier in batches of ten. Suppose that 52% of all such batches contain no defect

koban [17]

Answer:

P ( B0 / D0 ) = 0.59877

P ( B1 / D0 ) = 0.25793

P ( B2 / D0 ) = 0.14329

Step-by-step explanation:

Given:

- 0 be the event that the batch has 0 defectives = (0 ) = 0.52

- 1 be the event that the batch has 1 defectives = (1 ) = 0.28

- 2 be the event that the batch has 2 defectives = (2 ) = 0.2

- Two components are selected

Find:

What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions?

(a) Neither tested component is defective.

Solution:

Let 0 be the event that neither selected component is defective.

- The event 0 can happen in three different ways:

(i) Our batch of 10 is perfect, and we get no defectives in our sample of two;

P(i) = P(B0) = 0.52

(ii) Our batch of 10 has 1 defective, but our sample of two misses them;

P ( no defect / B1 ) = P ( no defect ) / P ( B 1 )

= 9C2 / 10C2 = 0.8

P ( ii ) = 0.28*0.8 = 0.224

(iii) Our batch has 2 defective, but our sample misses them.

P ( no defect / B2 ) = P ( no defect ) / P ( B 2 )

= 8C2 / 10C2 = 56/90

P ( iii ) = 0.2*56/90 = 0.124444

- Then,

P(Do) = P(i) + P(ii) + P(iii)

P(Do) = 0.52 + 0.224 + 0.124444 = 977/1125

We use the general conditional probability formula:

P ( B0 / D0 ) = P ( B0 & D0 ) / P( D0 )

P ( B0 / D0 ) = 0.52*1125 / 977 = 0.59877

P ( B1 / D0 ) = P ( B1 & D0 ) / P( D0 )

P ( B1 / D0 ) = 0.224*1125 / 977 = 0.25793

P ( B2 / D0 ) = P ( B2 & D0 ) / P( D0 )

P ( B2 / D0 ) = 0.12444*1125 / 977 = 0.14329

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3 years ago