Y + 2 = -3/4 ( x +1 ) in standard form

anygoal [31]

7/9 + 5/6 + 3 2/3
7/9 + 5/6 + 11/3
Common denominator is 18 so we get
14/18 + 15/18 + 66/18
= 95/18 cups of milk
Multiply this by 3 (since he uses 1/3 cups per latte)
= 285/18 = 15 15/18 = 15 5/6 cups of latte

7 0

3 years ago

(2n^3+20n^2+n)/10n^2

gizmo_the_mogwai [7]

The simplified form of this polynomial is (n(20n + 1))/5

3 0

3 years ago

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume t

motikmotik

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

$99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Step-by-step explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.4957$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where $\bar{x}$ is the sample mean and Margin of error is given by

$$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

$MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\$

So the required 99% confidence interval is

$\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

8 0

4 years ago

What is -1.5 in fraction form

Mrrafil [7]

-3/2 or -1 1/2 is the answer

6 0

3 years ago

I need help ASAP!! Please explain how to do solve the problem.

olga55 [171]

Answer:

3

Step-by-step explanation:

The co-ordinates of the smaller and larger triangles are 4,5 3,2 & 6,3 and 12,15 9,6 18,9

To find out dilation factor,

Co-ordinates of smaller triangles* dilation factor=co-ordinates of

larger triangles(or use (kx,ky) where k is the dilation factor and x and y are the co-ordinates)

(4,5) (3,2) (6,3)k=(12,15) (9,6) (18,9)

Equate and solve,

k=(12,15) (9,6) (18,9)/(4,5) (3,2) (6,3)[Informal]

k=3

6 0

3 years ago