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2 years ago

Identify the series as geometric or arithmetic. 2, 4/3, 8/9, 16/27

1 answer:

8 0

Hello,

$\dfrac{\frac{4}{3}}{2}=\dfrac{4}{6}=\dfrac{2}{3}$

$\dfrac{\frac{8}{9}}{\frac{4}{3}}=\dfrac{8*3}{9*4}=\dfrac{2}{3}$

$\dfrac{\frac{16}{27}}{\frac{8}{9}}=\dfrac{16*9}{27*8}=\dfrac{2}{3}$

The sequence is geometric .
(a serie is a sum)

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What is the slope of the line threw (-7,-2) and (-6,7)

miv72 [106K]

Answer:

Step-by-step explanation:

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$m=\frac{y_2-y_1}{x_2-x_1}$

Here the points are (-7,-2) and (-6,7). Substitute and simplify.

$m=\frac{7--2}{-6--7}=\frac{7+2}{-6+7}=\frac{9}{1}=9$

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

5,930,000 in scientific notation

Arturiano [62]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Mr. Vestas sells electricity from a wind turbine to his local utility. He receives 3¢ per kilowatt hour for the first 500 kilowa

Grace [21]

Answer:

Step-by-step explanation:

We are expected to solve the weighted average price per kilowatt hour.

Step one:

The formula is

W =∑wX/∑w

W=weighted average

w=weights applied to x values

X=data values to be averaged

Step two:

given the total killowatt hour is 1800

the 1st = 500 at 3¢

2nd= 1000 at 2¢

3rd = (1800-1000+500) at 1¢ =(1800-1500) at 1¢ = (300) at 1¢

Step two:

Substituting into the expression we have

W =(500*3)+(1000*2)+(300*1)/500+1000+300

W=1500+2000+300/1800

W=3800/1800

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The average price per kilowatt hour the utility paid to Mr. Vestas last month is 2.11¢

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2 years ago

Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. U

kogti [31]

Answer:

11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Forecast of rain.

Event B: Raining.

In recent years, it has rained only 5 days each year.

A year has 365 days. So

$P(B) = \frac{5}{365} = 0.0137$

When it actually rains, the weatherman correctly forecasts rain 90% of the time.

This means that $P(A|B) = 0.9$

Probability of forecast of rain:

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$P(A) = 0.0137*0.9 + 0.9863*0.1 = 0.11096$

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$P(B|A) = \frac{0.0137*0.9}{0.11096} = 0.1111$

11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

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2 years ago