One of the most important parts about creating great photographs starts with choosing what to photograph

5. Assume that you and your best friend ench have $1000 to invest. You invest your money

Bezzdna [24]

Correct question reads;

Assume that you and your best friend each have $1000 to invest. You invest your money in a fund that pays 10% per year compound interest. Your friend invests her money at a bank that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount for each of you is:

(a) You have $10 more than she does

(b) You have $100 more than she does

(c) You both have the same amount of money

(d) She has $10 more than you do

Answer:

(d) She has $10 more than you do

Explanation:

Using the compound interest formula

A= P [ (1-i)^n-1

Where P = Principal/invested amount, i = annual interest rate in percentage, and n = number of compounding periods.

My compound interest is:

= 1000 [ (1-0.1)^1-1

= $1000

$1,000 + $1,000 invested= $2,000 total amount received.

My friend's simple interest is;

To determine the total amount accrued we use the formula:

P(1 + rt) Where:

P = Invested Amount (1000)

I = Interest Amount (10,000)

r = Rate of Interest per year (10% or 0.2)

t = Time Period (1 )

= 1000 (1 + rt)

= 1000 (1 + 0.1x1)

= $1100 + $1000 invested = $2100 total amount received.

Therefore, we observe that she (my friend) has $100 more than I do.

5 0

2 years ago

Independent auto lots usually have ____ finance rates than dealerships

wel

Independent auto lots usually have higher finance rates than dealerships

Explanation:

The finance rates that are charged by the dealers are lower than the finance charges that are charged by the independent auto. In case if you are getting financed through dealerships, you can also negotiate with them to charge finance rates and lower the charges of the finance.

But this negotiation and lowering of the finance rates is not possible with the independent auto lots and thus they charge higher rates compared to the dealerships.

8 0

2 years ago

(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0

2 years ago

Are designed to make it easier for employees to get health and safety Information about

iren [92.7K]

Answer:

what the options

Explanation:

4 0

2 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .