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docker41 [41]
2 years ago
13

Mr. McKay wrote an algebra test with a total of 15 questions consisting of multiple choice and short response style questions. M

ultiple choice questions, x, were worth 5 points each and short response questions, y, were worth 10 points each. There were 100 total possible points on the test. Write and graph a system of linear equations for this situation to determine the number of each type of question.
Mathematics
1 answer:
tankabanditka [31]2 years ago
5 0

Step-by-step explanation:

x = number of multiple choice questions

y = number of short response questions

x + y = 15

5x + 10y = 100

=>

x + 2y = 20

let's subtract the first from the second equation :

x + 2y = 20

- x + y = 15

--------------------

0 y = 5

x + y = 15

x + 5 = 15

x = 10

to graph you need to consider both equations as linear functions. and you need to transform them into e.g. a slope intercept form : y = ax + b

a is the slope, b is the y- intercept.

x + y = 15

transforms to

y = -x + 15

this line goes e.g. through the points (0, 15) and (1, 14).

and

x + 2y = 20

transforms to

2y = -x + 20

y = -x/2 + 10

this line goes e.g through (0, 10) and (2, 9).

the crossing point of both lines is the solution and should therefore be the point (10, 5) as calculated above.

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The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.
Mama L [17]

Answer:

a) P(Y > 76) = 0.0122

b) i) P(both of them will be more than 76 inches tall) = 0.00015

   ii) P(Y > 76) = 0.0007

Step-by-step explanation:

Given - The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.

To find - (a) If a man is chosen at random from the population, find

                    the probability that he will be more than 76 inches tall.

              (b) If two men are chosen at random from the population, find

                    the probability that

                    (i) both of them will be more than 76 inches tall;

                    (ii) their mean height will be more than 76 inches.

Proof -

a)

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{S.D}) > \frac{( 76- mean)}{S.D})

                 = P(Z >  \frac{( 76- mean)}{S.D})

                 = P(Z > \frac{76 - 69.7}{2.8})

                 = P(Z > 2.25)

                 = 1 - P(Z  ≤ 2.25)

                 = 0.0122

⇒P(Y > 76) = 0.0122

b)

(i)

P(both of them will be more than 76 inches tall) = (0.0122)²

                                                                           = 0.00015

⇒P(both of them will be more than 76 inches tall) = 0.00015

(ii)

Given that,

Mean = 69.7,

\frac{S.D}{\sqrt{N} } = 1.979899,

Now,

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{\frac{S.D}{\sqrt{N} } })) > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- 69.7)}{1.979899 }))

                 = P(Z > 3.182)

                 = 1 - P(Z ≤ 3.182)

                 = 0.0007

⇒P(Y > 76) = 0.0007

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