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3 years ago

Which of the following represents the sum of the series? 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26

Mathematics

1 answer:

Alja [10]3 years ago

4 0

<h2>The sum of the given series = 176</h2>

Step-by-step explanation:

The given series:

$6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26$

Here, first term(a) = 6 and common difference(d) = 8 - 6 = 2

The given series are in AP.

Let the number of term = n

We know that,

The nth term of an AP

∴ $a_{n}=a+(n-1)d$

⇒ 6 + (n - 1) 2 = 26

⇒ (n - 1) 2 = 26 - 6 = 20

⇒ n - 1 = $\dfrac{20}{2} =10$

⇒ n = 10 + 1 = 11

∴ The sum of the given series = $\dfrac{n}{2}(a+a_{n})$

= $\dfrac{11}{2}(6+26)$

= $\dfrac{11}{2}(32)$

= 11 × 16 = 176

Thus, the sum of the given series = 176

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Answer:

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Step-by-step explanation:

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3 years ago

The sum of the areas of two circles is 80 pi

katovenus [111]

The area of a circle is A = πr^2. We let A1 And A2 the areas of the circles and r1 and r2 the radius of each, respectivley.

A1 + A2 = 80π
Substitute the formula for the area,
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From the statement, we know that r2=2(r1).

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3 years ago

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X 0 1 2

P(X) 42% 50% 8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c) P(win first game) = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X 0 1 2

P(X) 42% 50% 8%

P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value $\mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66$