What is 70/10 equivalent fraction?

Using technology, determine the quarterly payment on a 4 year loan of $16,231 at 5.1% compounded quarterly. round your answer to

Vesna [10]

The quarterly payment will be $1,127.86

<h3>What is compound interest?</h3>

Compound interest is the interest on a loan or deposit that levied on both the first principal and the collected interest from past periods.

The amortization formula is ...

$A=\dfrac{P\timesr}{(1-(1+r)^{-n}}$

where A is the payment amount, P is the principal amount, r is the interest rate per period, and n is the number of periods.

Here, we have

P=$16,231, r=0.051/4=0.01275, n=4·4=16.

So, the payment is ...

$A=\dfrac{16231\times 0.01275}{1-(1.01275)^{-16}}$

$A=$1127.86$

Hence the quarterly payment amount is $1,127.86.

To know more about Compound interest, follow

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8 0

3 years ago

Jada raised 60% of $20 goal. How much did she raise?

natima [27]

Answer:

$12

Step-by-step explanation:

setting up a proportion and putting 20 of 100 and 60 over x, we multiply 60 and 20 then divide one hundred from both sides leabing. We should have 100/x = 1200/100. Doing this gives us x = 12.

4 0

3 years ago

Read 2 more answers

What is an algerbric inequality for x is at most 30

Soloha48 [4]

I can't write the symbol, but it would be X<30 and the <would have a horizontal line under it so that it would mean x is less than or equal to 30

6 0

3 years ago

Read 2 more answers

In the given right triangle, find the missing length.<br> 10m<br> 24m

adell [148]

Answer:

10m

Step-by-step explanation:

mark me brainliest.

4 0

3 years ago

A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale protot

Olegator [25]

Answer:

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, $c_{m}$ = The drag coefficient of the prototype, $c_{p}$

The medium of the test for the model, $\rho_m$ = The medium of the test for the prototype, $\rho_p$

The drag force is given as follows;

$F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}$

We have;

$L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m$

Therefore;

$\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2$

$\dfrac{L_p}{L_m} =\dfrac{17}{1}$

$\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{c_p \times A_p \times \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}$

$\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3$

$\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3$ = (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ ≈ 0.0002.

5 0

3 years ago