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lesantik [10]
3 years ago
9

Trig: What is the rate of change of a function over a specified interval?

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
3 0
<span>Find the average </span>rate of change over<span> the </span>interval<span> 1 < x < 3. Solution: If the </span>interval<span> is 1 < x < 3, then you are examining the points (1,4) and (3,16).</span>
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Sequences and Series precalc multiple choice, will give brainliest.
Gala2k [10]

9514 1404 393

Answer:

  False

Step-by-step explanation:

The given formula is the "explicit" formula for the sequence.

The recursive formula would be ...

  a[1] = 160

  a[n] = 1/2·a[n-1] . . . . each term expressed in as a function of previous terms

The given statement is false.

8 0
2 years ago
An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. The ACME Manuf
iren [92.7K]

Answer:

a) P(ACME) = 0.7

b) P(ACME/D) = 0.5976

Step-by-step explanation:

Taking into account that ACME manufacturing company makes 70% of the ELTs, if a locator is randomly selected from the general population, the probability that it was made by ACME manufacturing Company is 0.7. So:

P(ACME) = 0.7

Then, the probability P(ACME/D) that a randomly selected locator was made by ACME given that the locator is defective is calculated as:

P(ACME/D) = P(ACME∩D)/P(D)

Where the probability that a locator is defective is:

P(D) = P(ACME∩D) + P(B. BUNNY∩D) + P(W. E. COYOTE∩D)

So, the probability P(ACME∩D) that a locator was made by ACME and is defective is:

P(ACME∩D) = 0.7*0.035 = 0.0245

Because 0.035 is the rate of defects in ACME

At the same way, P(B. BUNNY∩D) and P(W. E. COYOTE∩D) are equal to:

P(B. BUNNY∩D) = 0.25*0.05 = 0.0125

P(W. E. COYOTE∩D) = 0.05*0.08 = 0.004

Finally, P(D) and P(ACME/D) is equal to:

P(D) = 0.0245 + 0.0125 + 0.004 = 0.041

P(ACME/D) = 0.0245/0.041 = 0.5976

5 0
3 years ago
Help me please thank you
ser-zykov [4K]
There is nothing to help with...
3 0
3 years ago
Read 2 more answers
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
3 years ago
Marcus sorts his 85 baseball cards into stacks of 9 cards each.How many stacks of 9 cards can Marcus make?4th grade homework
laiz [17]
9 stacks of cards and 4 cards left over because when you divide 85 by 9 you get 9.4. So you'd multiply just 9 to get 81 and 4 cards leftover
7 0
3 years ago
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