Write the equation for the model below. Then solve for x.

If the division problem is 2.25 how many places should you move the decimal point in the dividend

Kazeer [188]

I think you move it up and you divide 225 divided by X

4 0

3 years ago

The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs.

gizmo_the_mogwai [7]

Answer:

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(920≤ x≤1730) = 0.7078

Step-by-step explanation:

Step(i):-

Given mean of the Population = 1100 lbs

Standard deviation of the Population = 300 lbs

Let 'X' be the random variable in Normal distribution

Let x₁ = 920

$Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6$

Let x₂ = 1730

$Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1$

Step(ii)

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)

= P(-0.6 ≤Z≤2.1)

= P(Z≤2.1) - P(Z≤-0.6)

= 0.5 + A(2.1) - (0.5 - A(-0.6)

= A(2.1) +A(0.6) (∵A(-0.6) = A(0.6)

= 0.4821 + 0.2257

= 0.7078

Conclusion:-

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(920≤ x≤1730) = 0.7078

5 0

3 years ago

The area is (2x + 15) square centimeters.

emmainna [20.7K]

Answer:

$Area\:shaded=\frac{3}{5} (2x+15)$

Step-by-step explanation:

Let $s$ be the shaded area and $u$ be the unshaded area, then we know that

(1). $\frac{s}{u}=\frac{3}{2}$

and

(2). $s+u=(2x+15)$

We solve for $u$ in the first equation and get:

$u=\frac{2}{3}s$

and put this into the second equation and get:

$s+\frac{2}{3}s=(2x+15)$

$s(1+\frac{2}{3} )=(2x+15)$

$s(\frac{5}{3} )=(2x+15)$

$\boxed{s=\frac{3}{5} (2x+15)}$

4 0

3 years ago

Create a list of 5 potential jobs that students of pharmacology can obtain.

aalyn [17]

Answer:

What can you do with a Pharmacology Degree?

Pharmacy InternshipStaff PharmacistPharmacist Manager. ...

Clinical Research AssociateRegulatory Affairs SpecialistRegulatory Affairs Manager. ...

Staff NurseRegistered Nurse SupervisorPatient Care Manager. ...

Research TechnicianEnvironmental ScientistToxicologist.

7 0

3 years ago

Read 2 more answers

Suppose the firm in this example considers a second product that has a unit profit of $5 and requires 2 hours of production time

user100 [1]

The question is incomplete, here is the complete question

Recall the production model from Section 1.3:

Max 10x

s.t. 5x ≤ 40

x ≥ 0

Suppose the firm in this example considers a second product that has a unit profit of $5 and requires 2 hours for each unit produced. Assume total production capacity remains 40 units. Use y as the number of units of product 2 produced. . Show the mathematical model when both products are considered simultaneously.

Answer:

Max Profit: 10x + 5y

5x + 2y ≤ 40

x ≥ 0, y ≥ 0

Explanation:

x= number of units of product 1 produced

y = number of units of product 2 produced

Since the first product, x, has a unit profit of $10 and Max1 is 10x

Second product, y, has a unit profit of $5, Max2 = 5y

The maximum profit when both products are considered simultaneously is 10x + 5y

Max Profit = 10x + 5y

Time required for each unit of x is 5hours

Therefore, time required for x units is 5x hours

Time required for each unit of y is 2hours

Therefore, time required for y units is 2y hours

Time required for the simultaneous production of both products is 5x + 2y

Since production capacity remains 40 units, 5x+2y ≤40

NB: The values of x and y cannot be negative

5 0

3 years ago