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2 years ago

9

A series contains 18 numbers and has a sum of 4,185. The last number of the series is 275. What is the first number?

1 answer:

dusya [7]2 years ago

3 0

The sum of arithmetic series is given by:
Sn=n/2(a1+an)
where:
n=number of terms
a1=first term
an=nth term
but
n=18, an=275, Sn=4185
plugging the values in the formula we get:
4185=18/2(a1+275)
simplifying this we get:
4185=9(a1+275)
dividing through by 9 we get:
465=a1+275
thus
a1=465-275
a1=190

Answer: first number is 190

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Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll

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Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

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The possible ways of getting a total of 7,

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

Number of favorable outcomes = 7

Formula for probability:

$Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So, the possibility of getting a total of 7 = $\frac{6}{36}=\frac{1}{6}$

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = $\frac{2}{36}$ = $\frac{1}{18}$

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = $\frac{28}{36}$ = $\frac{7}{9}$

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

$\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11$

$\frac{11}{3}+\frac{11}{3}-\frac{77}{9}$

$\frac{22}{3}-\frac{77}{9}$

$\frac{66-77}{9}$

$-\frac{11}{9}$

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Therefore the expected payoff for this game is -$1.22.

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The regular octagon has a perimeter of 122.4 cm. Which statements about the octagon are true? Select two options. The length of

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$P=A'\times 20\\\\P=13125\times 20\\\\P=262500$

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