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3 years ago

The population of a city for years since 2000 is shown below.

Mathematics

1 answer:

victus00 [196]3 years ago

3 0

Answer:

<h2>qqqqqqqqqqqqqqqqqqq</h2>

Step-by-step explanation:

ppp

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Question 5 (1 point)

Airida [17]

Answer:

y = 90

Step-by-step explanation:

If y varies jointly as x and z, then;

y = kxz

If y = 18 when x = 2 and z = 3

18 = k(2)(3)

18 = 6k

k = 18/6

k = 3

To get y when x = 5 and z = 6

y = kxz

y = 3(5)(6)

y = 15*6

y = 90

Hence the value of y is 90

3 0

2 years ago

What is the probability of drawing the compliment of a king or a

inna [77]

Answer:

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

Step-by-step explanation:

Step(i):-

Let 'S' be the sample space associated with the drawing of a card

n (S) = 52C₁ = 52

Let E₁ be the event of the card drawn being a king

$n( E_{1} ) = 4 _{C_{1} } = 4$

Let E₂ be the event of the card drawn being a queen

$n( E_{2} ) = 4 _{C_{1} } = 4$

But E₁ and E₂ are mutually exclusive events

since E₁ U E₂ is the event of drawing a king or a queen

step(ii):-

The probability of drawing of a king or a queen from a standard deck of playing cards

P( E₁ U E₂ ) = P(E₁) +P(E₂)

$= \frac{4}{52} + \frac{4}{52}$

P( E₁ U E₂ ) = $\frac{8}{52}$

step(iii):-

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards

$P(E_{1}UE_{2}) ^{-} = 1- P(E_{1} U E_{2} )$

$P(E_{1}UE_{2}) ^{-} = 1- \frac{8}{52}$

$P(E_{1}UE_{2}) ^{-} = \frac{52-8}{52} = \frac{44}{52} = 0.846$

Conclusion:-

The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846

5 0

3 years ago

What is the answer to y=-0.5(0)+2

Korolek [52]

The answer is: Y= 2.

3 0

3 years ago

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Hey can you please help me posted picture of question

nirvana33 [79]

(8x^2+2x+8)+(3x^3+7x+4)=
8x^2+2x+8+3x^3+7x+4
3x^3+8x^2+(2+7)x+12=
3x^3+8x^2+9x+12

Answer: Option A. 3x^3+8x^2+9x+12

8 0

3 years ago

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raketka [301]

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3 years ago