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frosja888 [35]
3 years ago
15

Arrange tho following in ascending order: 2/9, 1/3, 5/12, 1/6​

Mathematics
2 answers:
telo118 [61]3 years ago
5 0

Answer:

2/9=0.22

1÷3=0.33

5÷12=0.41

1÷6=0.16

Step-by-step explanation:

5/12

1/3

2/9

1/6

Sergeu [11.5K]3 years ago
4 0

Step-by-step explanation:

5/12, 2/9, 1/6, 1/2 thats the answer

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A school has 17 tables in the cafeteria. Each table seats 12 students. What is the
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Answer: 204 students

Step-by-step explanation: To find  out the greatest number of students that can fit at the 17 tables in the school cafeteria, we simply need to multiply the number of students that can fit at each table by the number of tables.

12 students can fit at 1 table and there are 17 tables.

So we have (12)(17) which is 204.

So the greatest number of students that can be seated din the cafeteria is 204 students.

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3 years ago
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Hunter installed tiles on a floor with a length of 12 feet and a width of 9 feet in 712 hours. how many square feet of tiles can
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5 0
3 years ago
Consider the triangle.
lbvjy [14]

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It is 26cm.

Step-by-step explanation:

Do  Pythagorean theorem, a^2+b^2=c^2

So 24^2+10^2,

576+100,

676.

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7 0
3 years ago
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A bird at the top of a tree looks down at a field mouse with an angle of depression of 65Á. If the field mouse is 30 meters from
Dafna11 [192]
Correct answer is 33.1 m.

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3 0
3 years ago
Consider a rabbit population​ P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squa
maria [59]

Solution:

Given :

$\frac{dP}{dt}= aP-bP^2$         .............(1)

where, B = aP = birth rate

            D = $bP^2$  =  death rate

Now initial population at t = 0, we have

$P_0$ = 220 ,  $B_0$ = 9 ,  $D_0$ = 15

Now equation (1) can be written as :

$ \frac{dP}{dt}=P(a-bP)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$    .................(2)

Now this equation is similar to the logistic differential equation which is ,

$\frac{dP}{dt}=kP(M-P)$

where M = limiting population / carrying capacity

This gives us M = a/b

Now we can find the value of a and b at t=0 and substitute for M

$a_0=\frac{B_0}{P_0}$    and     $b_0=\frac{D_0}{P_0^2}$

So, $M=\frac{B_0P_0}{D_0}$

          = $\frac{9 \times 220}{15}$

          = 132

Now from equation (2), we get the constants

k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

        = $\frac{3}{9680}$

The population P(t) from logistic equation is calculated by :

$P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$

$P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$

$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

As per question, P(t) = 110% of M

$\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$

$ 220-88e^{\frac{-99}{2420} t}=200$

$ e^{\frac{-99}{2420} t}=\frac{5}{22}$

Now taking natural logs on both the sides we get

t = 36.216

Number of months = 36.216

8 0
4 years ago
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