Ask question

Ask question

All categories

2 years ago

5

A stock has produced returns of 11.9 percent, 5.6 percent, 16.4 percent, and -4.2 percent over the past four years, respectively

. what is the geometric average return?

1 answer:

EleoNora [17]2 years ago

8 0

The average annual return is

$\displaystyle ((1+0.119)(1+0.056)(1+01.64)(1-0.042))^{\frac{1}{4}}-1=\sqrt[4]{1.317687706368}-1\\\\ \approx 7.14\%$

You might be interested in

Find the sum (3.2+4x)+(18.25+6x)=

tatiyna

Answer: -2.145

Step-by-step explanation:

(3.2+4x)+(18.25+6x)=

Simplifying

(3.2 + 4x) + (18.25 + 6x) = 0

Remove parenthesis around (3.2 + 4x)

3.2 + 4x + (18.25 + 6x) = 0

Remove parenthesis around (18.25 + 6x)

3.2 + 4x + 18.25 + 6x = 0

Reorder the terms:

3.2 + 18.25 + 4x + 6x = 0

Combine like terms: 3.2 + 18.25 = 21.45

21.45 + 4x + 6x = 0

Combine like terms: 4x + 6x = 10x

21.45 + 10x = 0

Solving

21.45 + 10x = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-21.45' to each side of the equation.

21.45 + -21.45 + 10x = 0 + -21.45

Combine like terms: 21.45 + -21.45 = 0

0 + 10x = 0 + -21.45

10x = 0 + -21.45

Combine like terms: 0 + -21.45 = -21.45

10x = -21.45

Divide each side by '10'.

x = -2.145

Simplifying

x = -2.145

8 0

2 years ago

Read 2 more answers

Manny is making 12 quarts of orange juice from concentrate and water. The number of quarts of water is 3 times the number of qua

trasher [3.6K]

Answer: There are 3 quarts of concentrate and 9 quarts of water.

Step-by-step explanation:

Since we have given that

Ratio of number of quarts of water to number of quarts of concentrate is

3:1

Number of quarts of orange juice from concentrate and water = 12 quarts

So, the number of quarts of concentrate would be

$\dfrac{1}{4}\times 12\\\\=\dfrac{12}{4}\\\\=3\ quarts$

the number of quarts of water would be

$\dfrac{3}{4}\times 12\\\\=3\times 3\\\\=9\ quarts$

Hence, there are 3 quarts of concentrate and 9 quarts of water.

8 0

2 years ago

A portfolio consisting of four stocks is expected to produce returns of minus9%, 11%, 13% and 17%, respectively, over the next f

anastassius [24]

Answer:

standard deviation of these expected returns = 0.0295 or 2.95%

Step-by-step explanation:

The detailed step is shown in the attachment

8 0

2 years ago

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 26% below the target pressure.

ZanzabumX [31]

Answer:

a) 22.94 psi

b) $5.93\times10^{-5}$

Step-by-step explanation:

a)The pressure at which will trigger a warning is

31 - 31*0.26 = 22.94 psi

b) The probability that that the TPMS will trigger warning at 22.94 psi, given that tire pressure has a normal distribution with average of 31 psi and standard deviation of 2 psi

$f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}$

where x = 22.94, $\mu = 31, \sigma = 2$

$f(22.94)={\frac {1}{2 {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {22.94-31}{2 }}\right)^{2}}$

$f(22.94)=0.2e^{-8.12} = 5.93\times10^{-5}$

6 0

2 years ago

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

pochemuha

Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

6 0

2 years ago