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3 years ago

There are 9 students in a class: 7 boys and 2 girls.

Mathematics

1 answer:

Nutka1998 [239]3 years ago

4 0

Answer:

The answer is 2 out of 3

Step-by-step explanation:

Since there are only 2 girls in the class. The probability of 2 girls in the same group is low. The maximum chance of the group being all boys is 2 out of 3. The minimum is 1 out of 3, because the girls can be olin different groups.

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0.2 ( 10 - 5c ) = 5c - 16

Mumz [18]

Answer: c = 3

Step-by-step explanation: Distribute 0.2 to (10-5c)

2-1c=5c-16

Then subtract 2 on both sides and subtract the 5c.

-1c=5c-18

-6c=-18

Then divide -6 to -18

-18/-6 = 3

8 0

3 years ago

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What is 2 5/7 + 9 6/21

Vaselesa [24]

Your answer is 4

Hope this helps.

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3 years ago

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As of December 31, 2020 $ 13,000. As of December 31, 2021 $ 25,000The enacted tax rate was 25% for 2020 and thereafter.What shou

blondinia [14]

Answer:

$3,000

Step-by-step explanation:

The computation of the balance in Kent deferred tax liability is shown below:

= {(December 31 2021) × (enacted tax rate)} - {(December 31 2020) × (Enacted tax rate)}

= {($25,000 × 25%) - ($13,000 × 25%)}

= $6,250 - $3,250

= $3,000

Along with the journal entry is also shown for better understanding

Income tax expense $41,500

To Deferred tax liability $3,000

To Income tax payable $38,250 ($153,000 × 25%)

(Being the income tax expense is recorded)

We debited the income tax expense as it increases the expenses account and on other side we credited the other two accounts as these are also increased

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3 years ago

M is the midpoint of HA. The coordinate of His

Scrat [10]

H=(3x+5,3y)
A=(x-1,-y)

Step-by-step explanation:

ANSWER:(2x+2,y)

7 0

3 years ago

The position function of a particle moving along a coordinate line is given, wheresis in feet andtis in seconds.

Sati [7]

Answer:

a) s = (4-t)/(t^2+4)^2, a(t) = (2t^3-24t)/(t^2+4)^3

b) s = 0.2ft, v = 0.12 ft/s, a = -0.176 ft/s^2

c) t = 2s

d) slowing down for t < 2, speeding up for t > 2

e) 0.327 ft

Step-by-step explanation:

The position function of a particle is given by:

$s(t)=\frac{t}{t^2+4},\ \ \ t\geq 0$ (1)

a) The velocity function is the derivative, in time, of the position function:

$v(t)=\frac{ds}{dt}=\frac{(1)(t^2+4)-t(2t)}{(t^2+4)^2}=\frac{4-t^2}{(t^2+4)^2}$ (2)

The acceleration is the derivative of the velocity:

$a(t)=\frac{dv}{dt}=\frac{(-2t)(t^2+4t)^2-(4-t^2)2(t^2+4)(2t)}{(t^2+4)^4}\\\\a(t)=\frac{(-2t)(t^2+4)-4t(4-t^2)}{(t^2+4)^3}=\frac{2t^3-24t}{(t^2+4)^3}$ (3)

b) For t = 1 you have:

$s(1)=\frac{1}{1+4}=0.2\ ft\\\\v(1)=\frac{4-1}{(1+4)^2}=0.12\frac{ft}{s}\\\\a(1)=\frac{2-24}{(1+4)^3}=-0.176\frac{ft}{s^2}$

c) The particle stops for v(t)=0. Then you equal equation (2) to zero ans solve the equation for t:

$v(t)=\frac{4-t^2}{(t^2+4)^2}=0\\\\4-t^2=0\\\\t=2$

For t = 2s the particle stops.

d) The second derivative evaluated in t=2 give us the concavity of the position function.

$\frac{d^2s}{dt^2}=a(2)=\frac{2(2)^3-24(2)}{(2^2+4)^3}=-0.062$

Then, the concavity of the position function is negative. For t=2 there is a maximum. Before t=2 the particle is slowing down and after t=2 the particle is speeding up.

e) Due to particle goes and come back. You first calculate s for t=2, then calculate for t=5.

$s(2)=\frac{2}{2^2+4}=0.25\ ft$

$s(5)=\frac{5}{5^2+4}=0.172\ ft$

The particle travels 0.25 in the first 2 seconds. In the following three second the particle comes back to the 0.172\ ft. Then, in the second trajectory the particle travels:

0.25 - 0.127 = 0.077 ft

The total distance is the sum of the distance of the two trajectories:

s_total = 0.25 ft + 0.077 ft = 0.327 ft

6 0

3 years ago