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3 years ago

The table shows the time it took a group of students to complete a puzzle

Mathematics

2 answers:

mash [69]3 years ago

8 0

Answer:

a. 17 b. 16/24

Step-by-step explanation:

Studentka2010 [4]3 years ago

5 0

Answer:

a. 17 b. 16/24

Step-by-step explanation:

a. Find the midpoints of each class interval which is:

7.5 x 2 = 15

12.5 x 9 = 112.5

17.5 x 5 = 87.5

22.5 x 5 = 112.5

27.5 x 3 = 82.5

Add all of these up to give 410

Add up the total of the frequency column which is 24

410 / 24 = 17.0833.....

Rounded to 2 significant figures, the answer is 17

b. Add together all the people under 20 which totals to 16 and put this over the total of 24 to give the fraction 16/24

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Which expression is equivalent to the following expression 3+13+3+13+3+13 : a) 3(13) b) 3(3+13) c)3×13 d) 3(3×13)

Alex Ar [27]

Answer:

The answer is B

Step-by-step explanation:

3+3+3+13+13+13

This is 3 times 3

This is also 3 times 13

Which would be 3(3+13)

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3 years ago

What is 6 divided by 131.4

zavuch27 [327]

Answer
0.045662100456621
6/131.4

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3 years ago

Read 2 more answers

Perimeter refers to the distance around 2-D figure and calculate perimeter, add the side lengths of the figure

viva [34]

Answer:

True

Step-by-step explanation:

The perimeter is the combined length of all the side lengths of a figure.

3 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

3 years ago

Evaluate p2m−(np+r) if m=−32 , n=2 , p=−8 , and r=4

harkovskaia [24]

When you evaluate the equation you plug in the numbers or replace the variables with the numbers it’s giving you.
-8(2)(-32) - 2(-8) + 4
-16(-32) - 16 + 4
512 + 20
= 532
Hope this helps!

8 0

3 years ago