Jim has grades of 84, 65, and 76 on three math tests. What grade must he obtain on the next test to have an average of exactly 80 for the four tests?. A. 87 . B. 95 . C. 85 . D. 92

Answers

Answer 1

Answer: An average for the four tests:
Avg = ( 84 + 65 + 76 + x ) / 4
80 = ( 84 + 65 + 76 + x ) / 4 /* 4 ( multiple both sides of the equation by 4 )
320 = 225 + x
x = 320 - 225
x = 95
Answer:
B ) 95

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The first term of an infinite geometric sequence is 18, while the third term is 8. There are two possible sequences. Find the sum of each sequence.

Answers

Answer:

Step-by-step explanation:

The sequence is geometric with

a = 18

Third term = 8.

Let the second term be x.

The infinite geometric sequence will be in the form

18,x,8,......

For geometric sequence, the ratio of two consecutive terms are equal. Hence, we have

$(x)/(18)=(8)/(x)\n\nx^2=144\nx=\pm12$

So, there are two possible sequences

18,-12,8,......

and

18,12,8,......

Thus, the common ratios are

$r=(-12)/(18),(12)/(18)\n\nr=(-2)/(3),(2)/(3)$

The sum of the infinite geometric for $r=(-2)/(3)$

$S=(a)/(1-r)\n\n(18)/(1-(-2)/(3))\n\nS=(18)/((5)/(3))\n\nS=(54)/(5)$

The sum of the infinite geometric for $r=(2)/(3)$

$S=(a)/(1-r)\n\n(18)/(1-(2)/(3))\n\nS=(18)/((1)/(3))\n\nS=54$

Hello,
Let'q the ratio
$u_(1) =18\n u_(2) =18*q\n u_(3) =18*q^2=8\ ==\ \textgreater \ q^2= (4)/(9) ==\ \textgreater \ q=\pm\ (2)/(3) \n1) if \ q= (2)/(3) \ then \ \sum_(i=0)^(\infty)\ 18* ((2)/(3) )^i= 18*(1)/(1+ (2)/(3) ) = (18*3)/(5)= 10.8\n\n2) if \ q= -(2)/(3) \ then \ \sum_(i=0)^(\infty)\ 18* (-(2)/(3) )^i= 18*(1)/(1- (2)/(3) ) = (18*3)/(1)= 54\n\n$

I need help asap plz

Answers

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Answers

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Answers

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Answers

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Answers

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Answer:

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Step-by-step explanation:

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