Manuel created a factor tree and wrote the prime factorization of 60 shown below.

Nate's recipe calls for 3 cups of flour, but all of his

Rainbow [258]

He should use 48 tablespoons of flour in his recipe.

6 0

2 years 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.

5 0

3 years ago

Two surveys were independently conducted to estimate a population mean, μ. denote the estimates and their standard errors by x1

lyudmila [28]

Hi there what you need is lagrange multipliers for constrained minimisation. It works like this,

V(X)=α2σ2X¯1+β2\sigma2X¯2

Now we want to minimise this subject to α+β=1 or α−β−1=0.

We proceed by writing a function of alpha and beta (the paramters you want to change to minimse the variance of X, but we also introduce another parameter that multiplies the sum to zero constraint. Thus we want to minimise

f(α,β,λ)=α2σ2X¯1+β2σ2X¯2+λ(\alpha−β−1).

We partially differentiate this function w.r.t each parameter and set each partial derivative equal to zero. This gives;

∂f∂α=2ασ2X¯1+λ=0

∂f∂β=2βσ2X¯2+λ=0

∂f∂λ=α+β−1=0

Setting the first two partial derivatives equal we get

α=βσ2X¯2σ2X¯1

Substituting 1−α into this expression for beta and re-arranging for alpha gives the result for alpha. Repeating the same steps but isolating beta gives the beta result.

Lagrange multipliers and constrained minimisation crop up often in stats problems. I hope this helps!And gosh that was a lot to type!xd

3 0

3 years ago

Can y’all please help me with those question

nasty-shy [4]

Step-by-step explanation:

The Fifteenth Amendment had the most long and short term effects. It helped the Republican Party to win in 1870, and this party brought extreme changes to the south. With the passing of this amendment, African Americans felt more empowered to participate in politics and hold office.

.

3 0

3 years ago

Read 2 more answers

Mark has $500 in his savings account. If he saves $25 each month, how many months will it take him to have $1000.

scoray [572]

Answer: 20, I believe

Step-by-step explanation: You already have 500 which is half of 1000. 500 divided by 25 equals 20.

8 0

3 years ago