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Ratling [72]
3 years ago
9

Can you please help me on 4(-4+m)=-6(m-4)

Mathematics
2 answers:
guapka [62]3 years ago
3 0

Answer:

Step-by-step explanation:

4(-4+m)=-6(m-4)

-16-m=-6m+24

-16=-5m+24

-40=-5m

/-5 = /-5

m=8

GarryVolchara [31]3 years ago
3 0

Answer:

m = 4

Step-by-step explanation:

Given

4(- 4 + m) = 6(m - 4) ← distribute parenthesis on both sides

- 16 + 4m = 6m - 24 ( subtract 6m from both sides )

- 16 - 2m = - 24 ( add 16 to both sides )

- 2m = - 8 ( divide both sides by - 2 )

m = 4

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Below are 36 sorted ages of an acting award winner. find the percentile corresponding to age 3838 using the method presented in
horrorfan [7]

Question:

Below are 36 sorted ages of an acting award winner. find the percentile corresponding to age 38 using the method presented in the textbook. 18 19 20 20 21 23 25 31 33 34 35 37 38 47 50 51 55 55 58 59 62 62 64 66 69 70 70 70 71 71 72 74 74 75 77 80 percentile of value 38 = ______ ​(round to the nearest integer as​ needed.)

Answer:

33 percentile

Step-by-step explanation:

Given the ages:

18, 19, 20, 20, 21, 23, 25, 31, 33, 34, 35, 37, 38, 47, 50, 51, 55, 55, 58, 59, 62, 62, 64, 66, 69, 70, 70, 70, 71, 71, 72, 74, 74, 75, 77, 80.

We are told to find the percentile corresponding to age 38.

To find the percentile corresponding to age 38 simply means to find the percentage of ages that are less than 38.

From the given values, the number of ages less than 38 are 12. Therefore, the percentile corresponding to age 38 would be:

(Ages less than 38/ total number of ages) * 100

= \frac{12}{36} * 100 = 33.333

33.333 ≈ 33

The percentile corresponding to age 38 = 33 percentile

5 0
3 years ago
Bella types at a constant rate of 42 words per minute. Is the number of words she can type proportional to the number of minutes
solmaris [256]
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If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.
Pie
Given that f^{(n)}(0)=(n+1)!, we have for f(x) the Taylor series expansion about 0 as

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Replace n+1 with n, so that the series is equivalent to

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}

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\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}

Recall that for |x|, we have

\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}

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f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}
\implies f(x)=\dfrac1{(1-x)^2}
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Answer:

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