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9966 [12]
4 years ago
9

Help me find the mean of this data set.

Mathematics
2 answers:
Lina20 [59]4 years ago
8 0

Answer:

Finding the mean is very simple, add them all together and divide that sum by the number of data points you have; in this case there are 45

Step-by-step explanation:

\frac{74+67+74.25+64+72+...}{45} = mean

Leviafan [203]4 years ago
3 0

Answer:

Add in all together then divide it by the number of numbers.

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Write the equation of this line in point-slope form and slope-
DaniilM [7]

Answer:

y = -1x + 2

Step-by-step explanation:

y = -1x + b

5 = -1(-3) + b

5 = 3 + b

2 = b

y = -1x + 2

3 0
3 years ago
Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0
Snowcat [4.5K]

Answer:

y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

Step-by-step explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} = e^{-cs}

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)}  − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 5s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 3s + 2s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[(s + 2)(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒Y(s) = \frac{3e^{-2s} }{(s + 2)(s + 3)} -  \frac{4e^{-5s} }{(s + 2)(s + 3)}

Now,

Let

\frac{1}{(s+2)(s+3)} = \frac{A}{s+2}  + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

\frac{1}{(s+2)(s+3)} = \frac{1}{s+2}  + \frac{-1}{s+3}

So,

Y(s) = 3e^{-2s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}]

⇒Y(s) = 3e^{-2s} \frac{1}{(s + 2)} -    3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ e^{-2(t-2)}  - e^{-5(t - 2)} ] - 4u₅(t) [ e^{-2(t-5)}  - e^{-5(t - 5)} ]

⇒y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

It is the required solution.

3 0
3 years ago
John has 6 pairs of pants and 3 shirts. How many possible outfits consisting of one shirt and
Step2247 [10]

Answer:

18

Step-by-step explanation:

3 shirts for each pants

4 0
3 years ago
Read 2 more answers
When a number is increased by 9 it is five times its opposite. Find the number.
Alla [95]

Let the unknown number be x.

Then the opposite of this unknown number will be -x.

It is given to us in the question that when this unknown number, x is increased by 9, or in other words when 9 is added to this unknown number, x, then this becomes 5 times it's opposite, which is -x.

Thus, from the given condition in the question, the equation we get is as:

x+9=-5x

We will solve this by isolating x as:

x+5x=-9

6x=-9

x=-\frac{3}{2} =-1.5

Thus the required number is -1.5

7 0
3 years ago
Help!<br> Ratios and Porportions
SOVA2 [1]
We learned that ratios are value comparisons, and proportions are equal ratios. Ratios can be written with colons or as fractions. So, to compare the number of girls to boys in a litter of puppies, we can write 2:4 or 2/4 to say that there are two girls to four boys.
7 0
3 years ago
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