What happens when you combine -3x+2x

What is 1/5 times 5/6 use a fraction modelto to help you solve.

lara [203]

Answer:

$\frac{1}{6}$

Step-by-step explanation:

4 0

1 year ago

My 4 closet friends have weekly allowance of $13, $17, $24, $30. What are the possible values of my weekly allowance if the medi

Nata [24]

Answer:

The median is 20.5

the average is 21

Step-by-step explanation:

5 0

2 years ago

-112=16e Solve for e

snow_tiger [21]

Answer: -7 = e

Step-by-step explanation: To solve this equation for e, we need to get e by itself on the right side of the equation. Since e is multiplied by 16, in order to get e by itself, we need to divide by 16 on the right side of the equation. If we divide by 16 on the right side of the equation, we must also divide by 16 on the left side of the equation.

On the right side of the equation the 16's cancel and we have e. On the right side of the equation -112 divided by 16 is -7. Remember that a negative divided by a positive is a negative.

So we have -7 = e which is the solution to our equation.

To check our solution, we can plug -7 in for e in the original equation. So we have -112 = 16 (-7) or -112 = -112 which is a true statement so our answer checks.

5 0

2 years ago

Vectors! Please help! This is the last question OF MY WHOLE SEMESTER! I've tried it so many times and can't get it right. :(

Feliz [49]

Pictures always help; see the one I've attached for reference.

Call the first vector (from HQ to the supply drop) u and the second vector (from supply drop to medics) v. We then want to find w, the vector from the medics to HQ, which corresponds to the vector -(u + v). (This is because u + v is the vector pointing from HQ to the medics; we're talking about the one in the opposite direction.)

Write the vectors in horizontal/vertical component form:

u = (125 km) (cos 235º x + sin 235º y) = (-71.70 x - 102.39 y) km

v = (75 km) (cos 110º x + sin 110º y) = (-25.65 x + 70.48 y) km

Why these angles?

"55 degrees south of west" is 235º; due west is 180º from the positive horizontal axis, and you add 55º to this
"20 degrees west of north" is 110º; due north is 90º, so add 20º to this

Add the vectors:

u + v = (-97.35 x - 31.92 y) km

Multiply by -1 to get the vector w:

w = -(u + v) = (97.35 x + 31.92 y) km

The distance covered by this vector is equal to its magnitude:

||w|| = √((97.35 km)^2 + (31.92 km)^2) = 102.45 km

The direction θ is given by

tanθ = (31.92 km)/(97.35 km) ==> θ = 18.15º

6 0

2 years ago

In a certain city, the daily consumption of water (in millions of liters) follows approximately a gamma distribution with α = 2

Ne4ueva [31]

Answer:

The probability that on any given day the water supply is inadequate $0.1991$

Step-by-step explanation:

Given

α = 2 and β = 3

As per Gamma distribution Function

$P(X>9)= 1-P(X\leq 9)\\$

Expanding the function and putting the given values, we get -

$[tex]1 - \int\limits^9_0 {f(x;2,3)} \, dx \\1- \int\limits^0_0 {\frac{1}{9}xe^{\frac{-x}{3} } \, dx\\\\= 1- \frac{1}{9} [x(-3e^{\frac{-x}{3}}) -\int\limits {(-3e^{\frac{-x}{3}})} \, dx]^9_0= 1- 1/9 [x(-3e^{\frac{-x}{3}}) -9e^{\frac{-x}{3}})} \, dx]^9_0\\1-((\frac{-1}{3} *9*e^({\frac{-9}{3}})- e^(\frac{-9}{3}))- ((\frac{-1}{3} *0*e^(\frac{-9}{3})-e^{\frac{-0}{3}})\\1-((-3e^{\frac{-9}{3} }-e^{\frac{-9}{3}}-(0-1))\\1-(1-4e^{-3})\\1-(1-0.1991)\\1-0.8009\\0.1991$

The probability that on any given day the water supply is inadequate $0.1991$

5 0

3 years ago