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DiKsa [7]
3 years ago
14

A bank loaned out ​$16000​, part of it at the rate of 7% per year and the rest at 15% per year. If the interest received in one

year totaled ​$1500​, how much was loaned at 7 % question mark How much of the ​$16000 did the bank loan out at 7 % question mark
Mathematics
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

  $11,250

Step-by-step explanation:

Normally, when I work "mixture" problems, I like to choose the variable so it represents the amount of the largest contributor. Here, that would be the amount invested at 15%.

Unfortunately, the question asks for the amount invested at 7%. Using a variable to represent that amount means that there will be negative numbers involved. Care must be taken in the arithmetic.

Let x represent the amount invested at 7%. Then 16000-x is the amount invested at 15%. The total interest for one year is ...

  0.07x + 0.15(16000 -x) = 1500 . . . . sum of interest amounts at each rate

  -0.08x +2400 = 1500  . . simplify

  -0.08x = -900 . . . . . . . . subtract 2400

  x = 11250 . . . . . . . . . . . . divide by -0.08

The bank loaned $11,250 at 7%.

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Answer:

a. θ = 30°

b. μ = √3 / 15 ≈ 0.115

Step-by-step explanation:

Draw a free body diagram for each scenario (see attached figure).  The body has four forces acting on it:

  • Weight pulling down
  • Normal force perpendicular to the incline
  • Applied force parallel up the incline
  • Friction force parallel to the incline

Remember that friction opposes the direction of motion.  So when the body is sliding up, friction points down the incline.  And when the body is sliding down, friction points up the incline.

Now apply Newton's second law to each scenario, first in the normal direction, then in the parallel direction.

For sliding up, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

P₁ − f − mg sin θ = 0

P₁ − Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₁ − mgμ cos θ − mg sin θ = 0

Now for sliding down, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

And sum of the forces parallel to the incline:

∑F = ma

P₂ + f − mg sin θ = 0

P₂ + Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₂ + mgμ cos θ − mg sin θ = 0

We know that P₁ = 6 kg.wt, P₂ = 4 kg.wt, and mg = 10 kg.wt.

So we have two equations and two unknowns (μ and θ):

P₁ − mgμ cos θ − mg sin θ = 0

P₂ + mgμ cos θ − mg sin θ = 0

Let's start by adding the equations together:

P₁ + P₂ − 2 mg sin θ = 0

P₁ + P₂ = 2 mg sin θ

sin θ = (P₁ + P₂) / (2 mg)

Plugging in the values:

sin θ = (6 + 4) / (2 × 10)

sin θ = 1/2

θ = 30°

Now we can plug this into either equation and find μ.

P₁ − mgμ cos θ − mg sin θ = 0

6 − (10 cos 30°) μ − 10 sin 30° = 0

6 − 5√3 μ − 5 = 0

1 = 5√3 μ

μ = √3 / 15

μ ≈ 0.115

6 0
3 years ago
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