Answer:
A. Advertising
Explanation:
Advertising is a paid form of non-personal communication targeted to an audience and usually employed by business men to promote their goods and services. The mass media which includes; radio, television, newspapers, e-mail are the means though which products can be advertised.
The description of what Xander wants to do which includes paying for the non-personal message which would be communicated through mass media, fit the description of advertising.
Answer:
The present value of terminal value is $ 863,689.48
Explanation:
Terminal value=Cash flows at third year*(1+g)/WACC-g
cash flows at the third year is $64,000
g is the growth rate of net cash flows which is 2% in perpetuity
WACC is 8%
Terminal value=$64,000*(1+2%)/(8%-2%)
=$64000*1.02/0.06
=$ 1,088,000.00
The present value of terminal=terminal value*discount factor in year 3
discount factor in year=1/(1+8%)^3=0.793832241
Present value of terminal cash flow=1,088,000.00 *0.79383224
=$ 863,689.48
Answer:
<u>86 payments approximately</u>
<em>Explanation</em>:
<u>First</u>;
Find the monthly average interest rate,
=7%/12
=0.0058333333
<u>Second</u>;
Add the monthly average interest rate to the monthly payment
= $175 + 0.0058333333
= $175.00583333 (average total monthly balance)
<u>Third</u>;
Divide final account balance by the average total monthly balance
= 15,000 / 175.00583333
=85.71 payments.
Answer:
41.49 approx 42 months
Explanation:
To calculate the number of months, we use the formula for loan
p = r(pv) / 1 - (1+r)-n
make n subject of the formula
p ( 1 - ( 1+r) ^-n) = r(pv)
p - p (1+r)^-n = r(pv)
p (1+r)^-n = p-r(pv)
(1+r)^-n = (p-r(pv)) / p
( 1+r)^n = p / (p-r(pv))
n In( 1+r) = In (p / (p-r(pv))
n = In ( p/ ( p - r(pv)) / In ( 1 +r)
n is the number of months, p is the payment per months
pv is the present value of 5000
substitute the values given into the equation
n = (In ( 150 / (150 - ( 0.129 / 12 × 5000)) / ( In ( 1 + ( 0.129 / 12) = 41.49 approx 42 months
The difference between loose and dense connective tissue is there is much more space between the fibers and cells in loose connective tissue than in dense connective tissue.
Option D
Explanation:
The two types of conjunctive tissue found in animals are the loose and compact connective tissues.
Connective tissue primarily serves soft bodies structurally. It also facilitates the provision of the epithelial tissue of nutrients and oxygen.
Elastic conjunctive tissue includes fibres that are elastic, whereas thick connective tissue comprises tightly structured fibres. Therefore the principal difference in the density of the extracellular matrix in each type of conjunctive tissue is between loose and dense binding tissue.
