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Ad libitum [116K]
3 years ago
6

The second term in a geometric series is 10, and the seventh term is 10,240. Find the sum of the first six terms.

Mathematics
1 answer:
just olya [345]3 years ago
5 0

Answer:

The sum of the first 6 terms is 3,412.5.

Step-by-step explanation:

The second term of the geometric series is given by:

a_{2}=a_{1}*r

Where a1 is the first term and r is the common ratio. The seventh term can be written as a function of the second term as follows:

a_{7}=a_{1}*r^{6} \\a_{7}=a_{2}*r^{5} \\10,240 = 10*r^{5}\\r=\sqrt[5]{1024} \\r = 4

The sum of "n" terms of a geometric series is given by:

a_{1} = \frac{10}{4} = 2.5\\S_{n}=a_{1}(\frac{r^{n}-1 }{r-1})\\S_{6}=2.5(\frac{4^{6}-1 }{4-1})\\S_{6}=3,412.5

The sum of the first 6 terms is 3,412.5.

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A swimming pool is open for 7 1/2 hours during the day the pool keeps one lifeguard shift is 1 1/16 hours long how many shifts a
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7 1/17 shifts

Step-by-step explanation:

That would be 7 1/2 divided by 1 1/16

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certain computer virus can damage any file with probability 35%, independently of other files. Suppose this virus enters a folde
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Answer:

0.623 is the probability that between 800 and 850 files get damaged.

Step-by-step explanation:

We are given the following information:

We treat virus can damage computer as a success.

P( virus can damage computer) = 35% = 0.35

The conditions for normal distribution are satisfied.

By normal approximation:

\mu = np = 2400(0.35) = 840\\\sigma = \sqrt{np(1-p)} = \sqrt{2400(0.35)(1-0.35)} = $$23.36

We have to evaluate probability that between 800 and 850 files get damaged.

P(800 \leq x \leq 850) = P(\displaystyle\frac{800 - 840}{23.36} \leq z \leq \displaystyle\frac{850-840}{23.36}) = P(-1.712 \leq z \leq 0)\\\\= P(z \leq 0.428) - P(z < -1.712)\\= 0.666 - 0.043 = 0.623 = 62.3\%

P(800 \leq x \leq 850) = 62.3\%

0.623 is the probability that between 800 and 850 files get damaged.

8 0
3 years ago
A farmer has a 100 ft by 200 ft rectangular field that he wants to increase by 15.5% by cultivating a strip of uniform width aro
Advocard [28]

Answer:

(a) The strip should be 5ft wide

(b) The strip around the outside field is 10ft wide.

Step-by-step explanation:

Given:

Length of the rectangular field, L= 200 ft

width of the rectangular field, w = 100 ft

Area of the rectangular field, A = 200ft x 100ft = 20000 ft^2

let the width of the strip = x

The strip around the outside field = 2x

If the field is increased by 15.5%

New area of the field = 1.155 x 20000 = 23,100 ft^2

The increase in area of the field = 3,100 ft

3,100 = New area of field - old area of the field

3100 = (200 + 2x)(100 + 2x) - 20000

3100 = 20000 + 400x 200x + 4x^2 - 20000

3100 = 600x + 4x^2

Divide through by 4

775 = 150x + x^2

x^2 + 150x - 775 = 0

Factorize

(x + 155)(x-5) = 0

x = 5 ft

The strip should be 5ft wide.

The strip around the outside field = 2 x 5 ft = 10 ft

Thus, the strip around the outside field is 10ft wide.

3 0
3 years ago
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