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Mashcka [7]
3 years ago
5

A. Find the mean median and mode of the data.

Mathematics
1 answer:
IrinaK [193]3 years ago
4 0
26 26 29 30 31 33 33 33 34 34 70

A.
Mean:30.9
Add all numbers up and divide by however many there are.
Median:32
Number in the middle spot
Mode:33
Number that occurs the most

B.
Mean:34.45
Median:33
Mode:33
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Write out the following number is words: 3,508
IRISSAK [1]
Three thousand five hundred eight
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Please answer in only a fraction please..
fenix001 [56]

Answer:

19/40

Step-by-step explanation:

1.9*(1/4) =0.475

0.475=19/40

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3 years ago
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Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be
Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L  grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C'      (C' = kC)

taking exponents of both sides, we have

L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt}      (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k}  - \frac{C"}{k} e^{kt}

When t = 0, A(0) = 0 (since the forest floor is initially clear)

A = \frac{L}{k}  - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{0}\\\frac{L}{k}  = \frac{C"}{k} \\C" = L

A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

when t = 0(at initial time), the initial value of D =

D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

4 0
3 years ago
Which graph represents the solution set for the inequality -8 < 8 + 4x < 28
valina [46]

Answer:

The solution set for the inequality is the interval (-4, 5)

Step-by-step explanation:

-8 < 8 + 4x < 28

Subtracting 8:

-8 -8 < 8 + 4x - 8 < 28 - 8

-16 < 4x < 20

Dividing by 4:

-16/4 < 4x/4 < 20/4

-4 < x < 5

x ∈ (-4, 5)

8 0
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I need help on this question 133^4/133^1
earnstyle [38]

Answer:

133^3 (or 2352637)

Step-by-step explanation:

\frac{133^4}{133^1}=133^{4-1}=133^3 (or 2352637 for exact value)

4 0
3 years ago
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