Answer:
At 6% $3,529.412 will be invested
At 11% $6,470.588 will be invested
Explanation:
Let x be the investment for 6% stock
And (10,000-x) is the investment it 11% stock
Let I be interest earned on both investments.
Using the formula
Principal(p)= Interest(I)*Rate(r)*Time(t)
p/RT= I
So considering both investments
x/(6%*1)= (10,000-x)/(11%*1)
x/0.06= (10,000-x)/0.11
Cross-multiply
0.11x= 0.06(10,000-x)
0.11x= 600- 0.06x
Rearranging
0.11x+ 0.06x= 600
0.17x= 600
x= 600/0.17= 3,529.412 amount invested at 6%
Amount invested at 11%= 10,000-3,529.412
= 6,470.588
The engineer's real income today in terms of constant 1950 dollars is $14,400.
<h3>What is the real income?</h3>
Real income ls nominal income less inflation rate. Inflation rate is when there is a persistent rise in the general price levels of a country.
Real income = nominal income - inflation
Inflation = (1 + 6.6) x $6000 = $45,600
Real income = $60,000 - $45,600 = $14,400
To learn more about real income, please check: brainly.com/question/6616964
Answer:
1,200 shares held at a cost basis of $37.50
Explanation:
Since there are 1,000 shares are purchased
and the stock dividend is 20%
So the number of shares after the dividend is
= 1,000 × (1 + dividend percentage)
= 1,000 × (1 + 0.20)
= 1,000 × 1.20
= 1.200
And, the price per share is
= $44 + $1
= $45
So, the cost basis would be
= $45 ÷ 1.20
= $37.50
hence, the tax status of the investment is 1,200 shares held for cost at $37.50 basis
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.