1+12-9+23409-1238 -12313+122321

Brainliest goes to whoever answers correctly

SSSSS [86.1K]

Answer:

The answer is C.

5 0

3 years ago

Read 2 more answers

Which of the following statements is true? a. Bonds have more risk than stocks, but offer more return. b. Stocks have more risk

Ipatiy [6.2K]

Answer:

Option: b is correct.

( Stocks have more risk than bonds, but offer more return).

Step-by-step explanation:

Bonds are debts while stocks are stakes of ownership in a company.

Bonds pay a fixed rate of interest, and guarantee principal payment at the end of the term, they're generally considered to be safer than stocks. That doesn't mean bonds are 100% safe.

<em>" Most investment professionals consider bonds a safe component of portfolios. They're supposed to provide the stability and certainty that stocks can't "</em>

<em>" In bond we have a fixed interest whereas in stock the rates could go much high "</em>

Hence, option b is correct. ( Stocks have more risk than bonds, but offer more return).

7 0

4 years ago

Emily spent $55 from her savings account on a new dress explain how to describe the change in emilys savings balance in two diff

mote1985 [20]

You can either express the decrease that happened in the account savings by integers. To do this you simply subtract 55$ from the initial amount in the account. This is expressed as:
initial amount - 55$

or, you can express the decrease in the form of percentage by calculating how much is 55$ from the total amount. This is expressed as:
(5/initial amount) x 100

8 0

3 years ago

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

$Ae^{rt}(r^2-11r+30)=0$

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

$y(t) = C_1e^{5t}+C_2e^{6t}$

8 0

3 years ago

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

garri49 [273]

Answer:

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0

3 years ago