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3 years ago

13

Determine the smallest degree of the Maclaurin polynomial required for the error in the approximation of the function at the ind

icated value of x to be less than 0.001. sin(0.4)

1 answer:

Olin [163]3 years ago

8 0

Answer: 3

Step-by-step explanation:

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A football team scored 3 touchdowns, 3 extra

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3x6+3x1+4x3

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3 years ago

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If y varies directly with x, and y is 14 when x is 2, what is the value of x when y is 35?

Ainat [17]

Im pretty sure it's 17.5 since 35 is odd.

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3 years ago

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What exponent or power should I put in this equation to make it true? 2.5 x 10® = 25,000 (PLZ HELP)

Over [174]

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4

Step-by-step explanation:

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3 years ago

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Prove that an = 4^n + 2(-1)^nis the solution to

olga nikolaevna [1]

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute $a_n=4^n+2(-1)^n$ in the equation $a_n=3a_{n-1}+4a_{n-2}$ the equality is true.

Let's substitute first in the right hand side:

$3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})$

Now we use the distributive laws. Also, note that $(-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n}$ (this also works when the power is n-2).

$=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}$

$=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}$

$=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n$

then the sequence solves the recurrence relation.

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3 years ago

A shipment of 11 printers contains 2 that are defective. Find the probability that a sample of size 2, drawn from the 11, will

svet-max [94.6K]

The required probability is $\frac{36}{55}$

<u>Solution:</u>

Given, a shipment of 11 printers contains 2 that are defective.

We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.

Now, we know that, $\text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}$

Probability for first draw to be non-defective $=\frac{11-2}{11}=\frac{9}{11}$

(total printers = 11; total defective printers = 2)

Probability for second draw to be non defective $=\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}$

(printers after first slot = 10; total defective printers = 2)

Then, total probability $=\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}$

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3 years ago