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3 years ago

Which ordered pairs are in the solution set of the system of linear inequalities?

Mathematics

1 answer:

never [62]3 years ago

8 0

Options:

(2, 2), (3, 1), (4, 2)

(2, 2), (3, –1), (4, 1)

(2, 2), (1, –2), (0, 2)

(2, 2), (1, 2), (2, 0)

Answer:

A. (2, 2), (3, 1), (4, 2)

Step-by-step explanation:

Given

$y > -\frac{1}{3}x + 2$

$y < 2x + 3$

Required

Solve for x and y

To solve this, we make use of graphical method (see attachment for graph)

All points that lie on the shaded region are true for the inequality

Next, we plot each of the given options on the graph

A. (2, 2), (3, 1), (4, 2)

All 3 points lie on the shaded region.

<em>Hence, (a) is true</em>

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crimeas [40]

Answer:

y = 1/2x

Step-by-step explanation:

x - 2y = -4

x = -4 + 2y

x + 4 = 2y

2y = x + 4

y = 1/2x + 2

y = 1/2x

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3 years ago

g Nationwide, a certain disease occurs only a small proportion of people (.000215). A certain town has 74,000 residents. Use the

Andrei [34K]

Answer:

The probability that the town has 30 or fewer residents with the illness = 0.00052.

Step-by-step explanation:

So, we have the following set of data or information or parameters given from the question above and they are; the number of people living in that particular society/community/town = 74,000 residents and the proportion of people that the diseases affected = .000215.

The first step to do is to determine the expected number of people with disease. Thus, the expected number of people with disease = 74,000 × .000215 = 15.91.

Hence, the probability that the town has 30 or fewer residents with the illness = 1.23 × 10^-7 × 15.91^30/ 2.65253 × 10^-32 = 0.00052.

Note the formula used in the calculating the probability that the town has 30 or fewer residents with the illness = e^-λ × λ^x/ x!

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3 years ago

A man is 32 years older than his son. Ten years ago he was three times as old as his son. find the present age of each?

nekit [7.7K]

The answer to this equation is 17

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3 years ago

Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can

Thepotemich [5.8K]

Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

6 0

3 years ago

Charlotte purchased a pool for $7680 using a six-month deferred payment plan with an interest rate of 20.45%. She did not make a

Georgia [21]

$7680:100%=$x:120.45%, 100x=7680*120.45, x=(7680*120.45)/100, x=$9250.56
Cost of the pool with interest rate is $9250.56.
3yr*12mo=36mo
$9250.56/36mo=256.96$/mo

<span> Charlotte's monthly payment will be $256.96.</span>

6 0

3 years ago