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3 years ago

10x = -71 - 11y3x = -7 + 11y

Mathematics

2 answers:

lyudmila [28]3 years ago

6 0

(-6,-1) is the answer to your question.

WINSTONCH [101]3 years ago

5 0

If we add theses 2 equations we eliminate the y terms ( -11y + 11y = 0):-

so adding we get

13x = -78

x = -6

Substitute x = -6 in the first equation:-

10(-6) = -71-11y

11y = -71 + 60 = -11

y = 1

So the solution to the system of equations is x = -6,y = 1

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The sum of two even consecutive integers is 46 find the intergers

Anarel [89]

Answer:

First number = 22

Second number = 24

Step-by-step explanation:

Let the first number = x

Let the second number = x + 2

According to the question ,

The sum of two even consecutive ingers = 46.

so,

x + x + 2 = 46

2x + 2 = 46

2x = 46 - 2

2x = 44

x = 44 / 2

x = 22

∴ FIRST NUMBER = X

= 22

SECOND NUMBER = X + 2

= 22 + 2

= 24.

4 0

4 years ago

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Len [333]

Answer:

x=1

Step-by-step explanation:

1+4=5

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Answer:

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Step-by-step explanation:

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3 years ago

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Andrei [34K]

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3 years ago

If A, B, and C are mutually exclusive events with P(A) = 0.2, P(B) = 0.3, and P(C) = 0.4, determine the following probabilities:

fiasKO [112]

a. The events are mutually exclusive, so

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\boxed{0.9}$

b. For the same reason,

$P(A\cap B\cap C)=\boxed0$

Another way to see this is to use the result from part (a), and the inclusion/exclusion principle:

$P(A\cap B\cap C)=P(A)+P(B)+P(C)-P(A\cup B\cup C)$

and the right side reduces to 0.

c. Same as (b),

$P(A\cap B)=\boxed0$

d. Same as (b),

$P((A\cup B)\cap C)=\boxed0$

We can also use the distributivity rule for unions and intersections to write

$P((A\cup B)\cap C)=P((A\cap C)\cup(B\cap C))=P(A\cap C)+P(B\cap C)=0$

e. If the $A'$ is the complement of $A$ , then by DeMorgan's law,

$P(A'\cap B'\cap C')=P(A\cup B\cup C)'=1-P(A\cup B\cup C)=\boxed{0.1}$

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3 years ago