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3 years ago

Hi I really need help with this assignment plz

Mathematics

2 answers:

brilliants [131]3 years ago

6 0

the first option is the correct one

Dahasolnce [82]3 years ago

6 0

13/18 = 0.72

11/14 = 0.78

So 13/18 < (less than) 11/14.

Hope this Helps!!

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There is a bag with 6 marbles. Three marbles are blue, 2 are red and 1 is green. What is the probability that you will randomly

Dominik [7]

Answer:

Mike is not correct.

Step-by-step explanation:

The probability is an "out of" statement. meaning out of all 6 marbles if 2 are red how what is the probability of choosing a red marble the answer is 2 out of 6 or 1/3 or 0.3333333etc. For green the probability is 1 out of 6 or 1/6 or 0.16666666etc

8 0

3 years ago

Using the following figure, match the angles in the first column with their corresponding congruent angles in the second column.

Rus_ich [418]

Answer:

∠6 = ∠8

∠2 = ∠6

Step-by-step explanation:

Lines 'p' and 'q' are the parallel lines and line 'l' is a transversal line intersecting these lines at two distinct points.

By the property of alternate interior angles,

∠3 = ∠5

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By the property of corresponding angles,

∠2 = ∠6

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3 years ago

1. The ratio of students in 8th grade that preferred math

Marta_Voda [28]

Answer:

a. 180

Step-by-step explanation:

$\frac{4x}{10}$ = $\frac{72}{1}$

cross-multiply:

4x = 720

x = 180

5 0

3 years ago

If the scale factor is 3/2, find the larger square missing side.A) 8 inch) 9 inch) 10 inch) 12 inch

ch4aika [34]

Answer:

B) 9

Step-by-step explanation:

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2 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