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1 year ago

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Kim solved the equation below by graphing a system of equations. log2(3x-1)= log4(x+8) What is the approximate solution to the e

quation?

2 answers:

dem82 [27]1 year ago

8 0

Answer:

The solution to the given equation is at point (1.353, 1.613).

Step-by-step explanation:

Given : Kim solved the equation below by graphing a system of equations.

$\log_2(3x-1)=\log_4(x+8)$

To find : What is the approximate solution to the equation?

Solution :

Let, $y_1=\log_2(3x-1)$

and $y_2=\log_4(x+8)$

Now, we plot these two equations.

The graph of $y_1=\log_2(3x-1)$ is shown with green line.

The graph of $y_2=\log_4(x+8)$ is shown with violet line.

The solution to this system will be their intersection point.

The intersection point of these graph is (1.353, 1.613)

Refer the attached graph below.

Therefore, The solution to the given equation is at point (1.353, 1.613).

kifflom [539]1 year ago

6 0

Answer:

(1.35, 1.6)

Step-by-step explanation:

A screenshot of the graph is attached.

Once we graph the equations $y=\log_2(3x-1)$ and $y=\log_4(x+8)$ , the solution to this system will be their intersection point.

Tracing the graph on the calculator, we see that the point of intersection is at about (1.35, 1.6).

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Katrina has the option of an 8-year non-subsidized student loan of $32,000 at an annual interest rate of 3.5% or an 8-year subsi

Fiesta28 [93]

Answer: katrina will pay less interest on the 8-year non subsidized student loan

The total interest paid on the non-subsidized loan is USD 8,959

The total interest paid on the subsidized loan is USD 11,520

Step-by-step explanation:

Subsidized loan at 4.5% for 8years will be 4.5/100 × 32000 x 1/12= USD120/month, USD 1440/year. For 8 years, USD11,520

For non subsidized student loan at 3.5% will be 3.5/100 x 32000x 1/12= USD 93.33/month, USD 1,119per year and USD 8,959 for 8 years

6 0

1 year ago

Multiplying StartFraction 3 Over StartRoot 17 EndRoot minus StartRoot 2 EndRoot EndFraction by which fraction will produce an eq

ziro4ka [17]

The fraction which produce an equivalent fraction with a rational denominator is $\left(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\right)$

Explanation:

The equation is $\frac{3}{\sqrt{17}-\sqrt{2}}$

To find the rational denominator, let us take conjugate of the denominator and multiply the conjugate with both numerator and denominator.

Rewriting the equation, we have,

$\frac{3}{\sqrt{17}-\sqrt{2}}\left(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\right)$

Multiplying, we get,

$\frac{3(\sqrt{17}+\sqrt{2})}{(\sqrt{17})^{2}-(\sqrt{2})^{2}}$

Simplifying the denominator, we get,

$\frac{3(\sqrt{17}+\sqrt{2})}{17-2}$

Subtracting, the values of denominator,

$\frac{3(\sqrt{17}+\sqrt{2})}{15}$

Dividing the numerator and denominator,

$\frac{\sqrt{17}+\sqrt{2}}{5}$

Hence, the denominator has become a rational denominator.

Thus, the fraction which produce an equivalent fraction with a rational denominator is $\left(\frac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}\right)$

6 0

1 year ago

Read 2 more answers

Ken grew 4/5 of an inch last year. sang grew3/8 of an inch . who grew more and by how much?

Licemer1 [7]

Sang grew more. and by 17/40 ( I could be wrong)

4 0

2 years ago

A hair salon in Cambridge, Massachusetts, reports that on seven randomly selected weekdays, the number of customers who visited

morpeh [17]

Answer:

a: 28 < µ < 34

Step-by-step explanation:

We need the mean, var, and standard deviation for the data set. See first attached photo for calculations for these...

We get a mean of 222/7 = 31.7143

and a sample standard deviation of: 4.3079

We can now construct our confidence interval. See the second attached photo for the construction steps.

They want a 90% confidence interval. Our sample size is 7, so since n < 30, we will use a t-score. Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)

The t-value is: 1.943

We rounded down to the nearest person in the interval because we don't want to over estimate. It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate. It's better to use 28 and underestimate a little when considering customer flow.

5 0

1 year ago

An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They

arlik [135]

Answer:

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

$s_2 =5.12399$ population sample deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$

Now we have everything in order to replace into formula (1):

$-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137$

$-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087$

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0

1 year ago