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3 years ago

Consider the fibonacci sequence below, which is the first term in the sequence to exceed 120?

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

3 0

Answer:

144

Step-by-step explanation:

The sequence is:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946

so you can see the first term larger than 120 is the 144.

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ahrayia [7]

4 is 18 because 6x3 is 18

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3 years ago

Read 2 more answers

Help me please quick its a test

True [87]

Answer:

When we think of World War I, images of the bloody, muddy Western Front are generally what come to mind. Scenes of frightened young men standing in knee-deep mud, awaiting the call to go "over the top", facing machine guns, barbed wire, mortars, bayonets, hand-to-hand battles, and more. We also think of the frustrations of all involved: the seemingly simple goal, the incomprehensible difficulty of just moving forward, and the staggering numbers of men killed. The stalemate on the Western Front lasted for four years, forcing the advancement of new technologies, bleeding the resources of the belligerent nations, and destroying the surrounding countryside. I've gathered photographs of the Great War from dozens of collections, some digitized for the first time, to try to tell the story of the conflict, those caught up in it, and how much it affected the world. This entry is part 2 of a 10-part series on World War I. This installment focuses on Early Years on the front, part II will focus more on the final year of trench warfare.

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Step-by-step explanation:

4 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

Coffee table is 20% odd the original price of $115. If there is 7.5% sales tax, what is the final price of the coffee table?

netineya [11]

100%-20%= 80%=0.8

115(0.8)= 92

92(1.075)= 98.9

The final price of the coffee table is $98.90.

Hope this helps!

3 0

3 years ago

A certain truck weighs 2 metric tons. A person weighing 60 kilograms gets in the truck, followed by a 70 kilogram person. How ma

Gala2k [10]

1 metric ton is 1000 kilograms
2 metric tons is 2000 kg.
60 kilograms + 70 kilograms = 130 kilograms
2000 kg. + 130 kg. = 2130 kg.
The truck weighs 2130 kilograms.

3 0

3 years ago