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3 years ago

15. A scale drawing of a rectangular field has a length of 4.5 inches and a width of

Mathematics

1 answer:

jeyben [28]3 years ago

7 0

Answer:30feet

Step-by-step explanation:

To find this

We have

Scale=length on drawing over actual length

0.5/10=1.5/x

0.05x=1.5

x=30feet

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Solve the equation algebraically: 10 1 X= 7 I I need an explanation too

Snowcat [4.5K]

Elaborate please for

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2 years ago

Someeee one?????????????????

Ad libitum [116K]

Answer:

Option B) $a_{n} = 2\cdot 4^{n-1}$

Step-by-step explanation:

The given geometric sequence is

2, 8, 32, 128,....

The general form of a geometric sequence is given by

$a_{n} = a_{1}\cdot r^{n-1}$

Where n is the nth term that we want to find out.

a₁ is the first term in the geometric sequence that is 2

r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

$r=\frac{8}{2} = 4$

You can try other consecutive numbers too, you will get the same common ratio

$r=\frac{32}{8} = 4$

$r=\frac{128}{32} = 4$

So the common ratio is 4 in this case.

Substitute the value of a₁ and r into the above general equation

$a_{n} = 2\cdot 4^{n-1}$

This is the general form of the given geometric sequence.

Therefore, the correct option is B

Note: Don't multiply the first term and common ratio otherwise you wont get correct results.

Verification:

$a_{n} = 2\cdot 4^{n-1}$

Lets find out the 2nd term

Substitute n = 2

$a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8$

Lets find out the 3rd term

Substitute n = 3

$a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32$

Lets find out the 4th term

Substitute n = 4

$a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128$

Lets find out the 5th term

Substitute n = 5

$a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512$

Hence, we are getting correct results!

6 0

3 years ago

The number of pizzas ordered on friday evenings between 5:30 and 6:30 at a pizza delivery location for the last 10 weeks is show

Ainat [17]

For a smoothing constant of 0.2

Time period – 1 2 3 4 5 6 7 8 9 10

Actual value – 46 55 39 42 63 54 55 61 52

Forecast – 58 55.6 55.48 52.18 50.15 52.72 52.97 53.38 54.90

Forecast error - -12 -.6 -16.48 – 10.12 12.85 1.28 2.03 7.62 -2.9
The mean square error is 84.12

The mean forecast for period 11 is 54.38
For a smoothing constant of 0.8

Time period – 1 2 3 4 5 6 7 8 9 10

Actual value – 46 55 39 42 63 54 55 61 52

Forecast – 58 48.40 53.68 41.94 41.99 58.80 54.96 54.99 59.80

Forecast error - -12 6.60 -14.68 0.06 21.01 -4.80 0.04 6.01 -7.80The mean square error is 107.17

The mean forecast for period 11 is 53.56

Based on the MSE, smoothing constant of .2 offers a better model since the mean forecast is much better compared to the 53.56 of the smoothing constant of 0.8.

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3 years ago

Please help this is due today please i really need help it’s due in 4 hours!!!

Anarel [89]

Answer:

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Step-by-step explanation:

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2 years ago

Which of the following statements is not true?

icang [17]

Answer:

-3.7 > -3.1

Step-by-step explanation:

Because -3.1 more close to 0 so is supposed to be bigger than -3.7

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2 years ago