2. What is the diameter of the circle below?

Can anyone pls help me to solve question 2 f and g and pls provide me a explanation I’m with that questions for three days

zvonat [6]

Answer:

f) a[n] = -(-2)^n +2^n

g) a[n] = (1/2)((-2)^-n +2^-n)

Step-by-step explanation:

Both of these problems are solved in the same way. The characteristic equation comes from ...

a[n] -k²·a[n-2] = 0

Using a[n] = r^n, we have ...

r^n -k²r^(n-2) = 0

r^(n-2)(r² -k²) = 0

r² -k² = 0

r = ±k

a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q

We find p and q from the initial conditions.

__

f) k² = 4, so k = 2.

a[0] = 0 = p + q

a[1] = 4 = -2p +2q

Dividing the second equation by 2 and adding the first, we have ...

2 = 2q

q = 1

p = -1

The solution is a[n] = -(-2)^n +2^n.

__

g) k² = 1/4, so k = 1/2.

a[0] = 1 = p + q

a[1] = 0 = -p/2 +q/2

Multiplying the first equation by 1/2 and adding the second, we get ...

1/2 = q

p = 1 -q = 1/2

Using k = 2^-1, we can write the solution as follows.

The solution is a[n] = (1/2)((-2)^-n +2^-n).

5 0

3 years ago

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e

professor190 [17]

Answer:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

$n=15$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =16$ represent the sample variance

$\sigma^2_0 =25$ represent the value that we want to verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 25$

Alternative hypothesis: $\sigma^2$

The statistic is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And replacing we got:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0

3 years ago

What are the potential solutions to the equation 2ln(x+3)=0

skad [1K]

$2\ln(x+3)=0\implies \ln(x+3)^2=0\implies e^{\ln(x+3)^2}=e^0\implies (x+3)^2=1$

Expanding the left side gives

$x^2+6x+9=1\implies x^2+6x+8=0\implies (x+4)(x+2)=0$

which gives two solutions, $x=-4$ and $x=-2$ . But if $x=-4$ , then $\ln(x+3)=\ln(-1)$ , but this number isn't real, so $x=-4$ is an extraneous solution. Meanwhile if $x=-2$ , you get $\ln(-2+3)=\ln1=0$ , so this solution is correct.

"Potential solutions" might refer to both possibilities, but there is only one actual (real) solution.

4 0

4 years ago

Read 2 more answers

Yes or no please, question in picture

Nataly [62]

Answer:

yes

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Stella can type 150 words in 5 minutes. What is her rate per minute?