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Yuki888 [10]
3 years ago
7

On a graph where the y-axis and x-axis jump from 1 to 10 to 100, with approximately equal space between 1-10 and 10-100. Which o

f the following is/are reasonable interpretations of this graph?
a. the variables on the axes are shown using a logarithmic scale
b. large increases in island area have relatively small effects on species numbers
c. the difference in the number of species on islands measuring 1 square mile, contrasted with islands measuring 100 square miles IS THE SAME as the difference in number of species on islands measuring 10 square miles, contrasted with islands measuring 100 square miles
d. all of these options are valid interpretations of this graph
Mathematics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

d-all of these options are valid interpretations of this graph

Step-by-step explanation:

Logarithmic scales are not linear, and they are used for large range of possitive numbers. The value represented by each equidistant mark on the scale is the value at the previous mark multiplied by a constant, in this case 10.

So, in large islands, large incresaes have a small effect ( the distance is the same from islands thas has 100 to 90 square miles, than for 2 islands measuring 10 to 9).

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Step-by-step explanation:

5 times 5 is 25

8 times 5 is 40

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Step-by-step explanation:

Theoretically, there is not square root of neither 13 nor a negative number so the special symbol is used to represent the square root of a number.

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37.1 pt= how Many gallons? Type a whole number or decimal. Round to the appropriate number of significant digits.
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3 0
2 years ago
The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

6 0
2 years ago
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