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3 years ago

Find the sum of the arithmetic series given a1=8,a14=99,n=14.

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

8 0

Answer:

The sum of the series is 749

Step-by-step explanation:

Given

$a_1 = 8$

$a_{14} = 99$

$n = 14$

Required

The sum of the series

This is calculated using:

$S_n = \frac{n}{2}[a_1 + a_n]$

Substitute 14 for n

$S_{14} = \frac{14}{2}[a_1 + a_{14}]$

$S_{14} = 7[a_1 + a_{14}]$

Substitute values for a1 and a14

$S_{14} = 7[8 + 99]$

$S_{14} = 7*107$

$S_{14} = 749$

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Whats f? 0.62(f + –6) + 6.96 = 8.2

Semmy [17]

Answer:

f=8

Step-by-step explanation:

0.62(f+(-6)+6.96=8.2

0.62(f-6)+6.96=8.2

multiply both sides by 100

0.62(f-6)*100+6.96*100=8.2*100

Refine

62(f-6)+696=820

Subtract 696 from both sides

62(f-6)+696-696=820-696

Simplify

62(f-6)=124

Divide both sides by 62

62(f-6)/62 = 124/62

Simplify

f-6=2

add 6 to both sides

f-6+6=2+6

simplify

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3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

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Semenov [28]

Ratio of apples to strawberries in Angie's salad

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= 36:45 (because we multiplied both numerator and denominator by 9, to get the same denominator as the other ratio)

Ratio of apples to strawberries in Salim's salad

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= 35:45 (because we multiplied both numerator and denominator by 5, to get the same denominator as the other ratio)

36:45 ≠ 35:45

So, ratio of apples to strawberries in Angie's salad ≠ ratio of apples to strawberries in Salim's salad.

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So, ratio of apples to strawberries in Angie's salad < ratio of apples to strawberries in Salim's salad.

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