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vovikov84 [41]
3 years ago
12

What is the sum of an 8-term geometric series if the first term is -11, the last term 180,224, and the common ratio is -4? answe

rs to choose from. can somebody please solve this?
a)-143,231
b)-36,047
c)144,177
d)716,144
Mathematics
1 answer:
Blababa [14]3 years ago
4 0
The general formula for the sum of the n terms of a geometric progression is:

Sn = A1 (1 - r^n) / (1 - r)

In this case, n = 8; A1 = - 11, r = -4

S8 = -11 (1 - (-4)^8) / (1 -(-4)) = 144,177

Answer: option c.
 
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The answer you're looking for is 2.00

Step-by-step explanation:

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2 years ago
The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation St
saveliy_v [14]

Answer:

a)0.067

b)0.111

c)0.612

d)$687.28

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $385

Standard Deviation, σ = $110

We are given that the distribution of domestic airfares is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(domestic airfare is $550 or more)

P(x > 550)

P( x > 550) = P( z > \displaystyle\frac{550 - 385}{110}) = P(z > 1.5)

= 1 - P(z \leq 1.5)

Calculation the value from standard normal z table, we have,  

P(x > 550) = 1 - 0.933 = 0.067= 6.7\%

b) P(domestic airfare is $250 or less)

P(x \leq 250) = P(z \leq \displaystyle\frac{250-385}{110}) = P(z \leq -1.22)

Calculating the value from the standard normal table we have,

P( x \leq 250) = 0.111 = 11.1\%

c))P(domestic airfare is between $300 and $500)

P(300 \leq x \leq 500) = P(\displaystyle\frac{300 - 385}{110} \leq z \leq \displaystyle\frac{500-385}{110}) = P(-0.77 \leq z \leq 1.04)\\\\= P(z \leq 1.04) - P(z < -0.77)\\= 0.851 - 0.239 = 0.612 = 61.2\%

P(300 \leq x \leq 500) = 61.2\%

d) P(X=x) = 0.03

We have to find the value of x such that the probability is 0.03.

P(X > x)

P( X > x) = P( z > \displaystyle\frac{x - 385}{110})=0.03

= 1 -P( z \leq \displaystyle\frac{x - 385}{110})=0.03

=P( z \leq \displaystyle\frac{x - 385}{110})=0.997

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 385}{110} = 2.748\\x = 687.28

Hence, the domestic fares must be $687.28 or greater for them to lie in the highest 3%.

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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