Write an expression for cos 68° using sine

23^0 is equal to _____. 0 1 1/23 23

Aliun [14]

Answer:

1

Step-by-step explanation:

Anything to the power is 0 is equal to 1.

5 0

2 years ago

Read 2 more answers

Suppose students' ages follow a skewed right distribution with a mean of 24 years old and a standard deviation of 3 years. If we

natta225 [31]

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Mean of students' age = 24 years

Standard deviation of students' age = 3 years

Sample size = number of students = 350

So, according to options,

a. The shape of the sampling distribution is approximately normal.

It is true as n >30, we will use normal.

b. The mean of the sampling distribution is approximately 24-years old.

It is true as it is given.

c. The standard deviation of the sampling distribution is equal to 5 years.

It is not true as it is given 3 years.

Hence, Option 'c' is correct.

8 0

2 years ago

Simplify the expression 3x(x – 12x) + 3x2 – 2(x – 2)2. Which statements are true about the process and simplified product? Check

olasank [31]

We have the expression:
3x(x-12x) + 3x^2 - 2(x-2)^2

First, we will expand the power 2 bracket as follows:
3x(x-12x) + 3x^2 - 2(x^2 - 4x +4)

Then, we will get rid of the brackets as follows:
3x^2 - 36x^2 + 3x^2 - 2x^2 + 8x - 8

Now, we will gather the like terms and add them as follows:
-32 x^2 + 8x - 8

We can take the 8 as a common factor:
8 ( -4x^2 + x -1)

3 0

2 years ago

A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s

denpristay [2]

Answer:

For this case the 95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=40-1=39$

For this case the 95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

3 0

2 years ago

Find the equation of a line parallel to -3x-5y=4 that contains the point (4,3). Write the equation and slope intercept form

allsm [11]

Answer:

y= 10

Step-by-step explanation:

-3x-5y=4

+3x +3x

-5y=7x

+5y +5y

y= 10

4 0

2 years ago