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3 years ago

Solve for n 3n-2=2n N=

Mathematics

2 answers:

krek1111 [17]3 years ago

7 0

Answer:

Hey there!

3n-2=2n

n-2=0

n=2

Let me know if this helps :)

DENIUS [597]3 years ago

6 0

3n - 2 = 2n

n =?

3n - 2n = 2

n = 2 ✔️

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Fynjy0 [20]

Total of all Gross Wages minus all pre- tax deductions. Total of all Pre-Tax and. Post-Tax deductions. For this pay period.

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2 years ago

Simplify the expression. (p^9)^−2

EastWind [94]

p9/p3 Final result : p6 Reformatting the input :

Changes made to your input should not affect the solution:

(1): "p3" was replaced by "p^3". 1 more similar replacement(s).

Step by step solution :Step 1 : p9 Simplify —— p3 Dividing exponential expressions :

1.1 p9 divided by p3 = p(9 - 3) = p6

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3 years ago

Read 2 more answers

Unknown to a medical researcher, 6 out of 25 patients have a heart problem that will result in death if they receive the test dr

Fiesta28 [93]

Using the hypergeometric distribution, it is found that there is a 0.0002 = 0.02% probability that exactly 6 patients will die.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

The values of the parameters for this problem are:

N = 25, k = 6, n = 8.

The probability that exactly 6 patients will die is P(X = 6), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 6) = h(6,25,8,6) = \frac{C_{6,6}C_{19,2}}{C_{25,8}} = 0.0002$

0.0002 = 0.02% probability that exactly 6 patients will die.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

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1 year ago

A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s

mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, $\theta=25$ °

Initial speed is, $u=15\ m/s$

Time of flight is, $t=3.0\ s$

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

$u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s$

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = $-h$ (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, $g=-9.8\ m/s^2$

Now, using the following equation of motion;

$-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m$

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, $v_y=0\ m/s$

Now, using the following equation of motion, we have:

$v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m$

Therefore, the maximum height of the brick is 2.05 m.

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2 years ago

If a(9,18) and b(1,12) find

galina1969 [7]

$AB= \sqrt{(9-1)^2+(18-12)^2} = \sqrt{8^2+6^2}= \sqrt{64+36} = \sqrt{100}=10$

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3 years ago