Answer:
15
Step-by-step explanation:
the diagonal of a rhombus are perpendicular, so we can use the Pythagorean theorem to solve this problem:
CL = √17^2 - 8^2 = √289 - 64 = √225 = 15
Using the t-distribution, it is found that:
a. The <u>margin of error</u> is of 4.7 homes.
b. The 98% confidence interval for the population mean is (19.3, 28.7).
The information given in the text is:
- Sample mean of
. - Sample standard deviation of
. - Sample size of
.
We are given the <u>standard deviation for the sample</u>, which is why the t-distribution is used to solve this question.
The confidence interval is:
![\overline{x} \pm M](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%5Cpm%20M)
The margin of error is:
![M = t\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20t%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
Item a:
The critical value, using a t-distribution calculator, for a two-tailed <u>98% confidence interval</u>, with 23 - 1 = <u>22 df</u>, is t = 2.508.
Then, the <em>margin of error</em> is:
![M = t\frac{s}{\sqrt{n}} = 2.508\frac{9}{\sqrt{23}} = 4.7](https://tex.z-dn.net/?f=M%20%3D%20t%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%202.508%5Cfrac%7B9%7D%7B%5Csqrt%7B23%7D%7D%20%3D%204.7)
Item b:
The interval is:
![\overline{x} - M = 24 - 4.7 = 19.3](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20-%20M%20%3D%2024%20-%204.7%20%3D%2019.3)
![\overline{x} + M = 24 + 4.7 = 28.7](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%2B%20M%20%3D%2024%20%2B%204.7%20%3D%2028.7)
The 98% confidence interval for the population mean is (19.3, 28.7).
A similar problem is given at brainly.com/question/15180581
Answer:
b
Step-by-step explanation:
did this help
The true statements are:
- Tyson's expression is not equivalent to the original expression
- The equivalent expression is:
![\frac{1}{2}g- \frac 1{5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dg-%20%5Cfrac%201%7B5%7D)
<h3>What are equivalent expressions?</h3>
Equivalent expressions are expressions that have equal values
The original expression is given as:
![\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}](https://tex.z-dn.net/?f=%5Cfrac%2015g-%20%5Cfrac%201%7B10%7D-%20g%20%2B%201%20%5Cfrac%7B3%7D%7B10%7Dg%20-%5Cfrac%7B1%7D%7B10%7D)
Collect like terms
![\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10} -\frac{1}{10}](https://tex.z-dn.net/?f=%5Cfrac%2015g-%20%5Cfrac%201%7B10%7D-%20g%20%2B%201%20%5Cfrac%7B3%7D%7B10%7Dg%20-%5Cfrac%7B1%7D%7B10%7D%20%3D%20%5Cfrac%2015g%20-%20g%20%2B%201%20%5Cfrac%7B3%7D%7B10%7Dg-%20%5Cfrac%201%7B10%7D%20%20-%5Cfrac%7B1%7D%7B10%7D)
Evaluate the like terms
![\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}](https://tex.z-dn.net/?f=%5Cfrac%2015g-%20%5Cfrac%201%7B10%7D-%20g%20%2B%201%20%5Cfrac%7B3%7D%7B10%7Dg%20-%5Cfrac%7B1%7D%7B10%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dg-%20%5Cfrac%201%7B5%7D)
Tyler's equivalent expression is given as:
![\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}](https://tex.z-dn.net/?f=%5Cfrac15g-g%2B%201%20%5Cfrac3%7B10%7Dg-%5Cfrac%201%7B10%7D-%5Cfrac%2045g%2B1%20%5Cfrac%7B1%7D%7B10%7D)
Collect like terms
![\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}](https://tex.z-dn.net/?f=%5Cfrac15g-g%2B%201%20%5Cfrac3%7B10%7Dg-%5Cfrac%201%7B10%7D-%5Cfrac%2045g%2B1%20%5Cfrac%7B1%7D%7B10%7D%20%3D%20%5Cfrac15g-g%2B%201%20%5Cfrac3%7B10%7Dg-%5Cfrac%2045g-%5Cfrac%201%7B10%7D%2B1%20%5Cfrac%7B1%7D%7B10%7D)
Evaluate the like terms
![\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g](https://tex.z-dn.net/?f=%5Cfrac15g-g%2B%201%20%5Cfrac3%7B10%7Dg-%5Cfrac%201%7B10%7D-%5Cfrac%2045g%2B1%20%5Cfrac%7B1%7D%7B10%7D%20%3D%20-%5Cfrac%203%7B10%7Dg)
The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.
Hence, Tyson's expression is not equivalent to the original expression
Read more about equivalent expressions at:
brainly.com/question/9603710
I don’t know the answer but I hope someone else doesnt