HELP PLS PLS PLS PLS

ruslelena [56]

$i=\sqrt{-1}\to i^2=-1\\\\5^3=5\cdot5\cdot5=125\\\\i^9=i^8\cdot i=(i^2)^4i=(-1)^4i=1i=i\\\\\boxed{5^3i^8=125i}$

7 0

3 years ago

2) Portia deposited $16,879.35 into a savings account paying 9.6% interest compounded annually for 3 years. How much money will

Monica [59]

Answer:

is 20,000

Step-by-step explanation:

0hhggyuibbhu88j b

7 0

3 years ago

Use the graph. Which equation is true from the information in the graph?

Nikolay [14]

From the graph, we see that two similar triangles are created by using the slope of the line.
• Because the two triangles appear to have congruent angles and proportional side lengths, we can conclude with the information we have that the two triangles are indeed similar. They are the same shape but proportional with congruent angles.
• Therefore, a/b = c/d because the scale factor remains the same of both ratios.
•We can also see that in larger triangle, the slope = 3/6 and the slope of the smaller one = 2/4. 3/6 = 2/4 because 3/6 = 1/2 and 2/4 = 1/2. Therefore, the slopes are proportional and equal.
• Because the slopes are proportional and the triangles are proportional, the a/b = c/d.

7 0

3 years ago

PLEASE HELP i’m desperate

gulaghasi [49]

Answer:

id : 237 514 2470

pwd : frmCV5

4 0

3 years ago

The national average for the number of students per teacher for all U.S. public schools in 15.9. A random sample of 12 school di

Nady [450]

Answer:

The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 12 - 1 = 11

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.201

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.201\frac{\sqrt{4.41}}{\sqrt{12}} = 1.3$

In which s is the standard deviation of the sample(square root of the variance) and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 19.2 - 1.3 = 17.9 students.

The upper end of the interval is the sample mean added to M. So it is 19.2 + 1.3 = 20.5 students

The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.

5 0

3 years ago