IS DUE PLZ HELP Mabel ate break fast at a restaurant. The bil lcame to $38. If she left an 18% tip, how much was the tip?

Which table of values represents the equation Y= 2X +5?

barxatty [35]

Answer:

12

Step-by-step explanation:

212113

8 0

3 years ago

The Kroger Company is one of the largest grocery retailers in the United States with over two thousand grocery stores across the

Ilia_Sergeevich [38]

Answer:

1) Categorical

2) Norminal

Step-by-step explanation:

1)

The data collected by Kroger in this example categorical or quantitative identified below:

From the given information, Kroger uses an online customer opinion to obtain the data about its products and services. All the questions based on yes or no type questions. Here the questions are ‘products that have a brand name, products that are environmentally friendly, products that are organic’ these type of questions cannot be expressed numerically so the data collected by Kroger Company is categorical variable because these answers of the questions cannot be counted.

Any variable which is grouped into two or more attributes then it is a categorical variable. The data collected by Kroger Company is categorical variable and any variable which can be counted or measured in numerical then it is quantitative variable.

2)

The measurement scale is identified below:

Here the variable cannot be counted in numerical sale so the level of measurement cannot be ratio, interval because ratio and interval scale can be used for numerical data. The nominal scale can be used to identify the ‘products that have a brand name, products that are environmentally friendly, products that are organic, products that have been recommended by others’ because natural order need not be used.

The ratio and interval scale can be used for Quantitative data and nominal and ordinal scale can be used for Qualitative data. When the order is needed to categorize the objects Ordinal scale is used, when the order is not needed to categorize the objects Nominal scale is used.

3 0

3 years ago

Please solve this:<br> 6-2x=3

Westkost [7]

Answer:

-4

Step-by-step explanation:

6 - 2x = 3

-2x = 3 + 6

-2x = 8

x= $\frac{8}{-2}$

x= -4

I apologize if this isn't the right answer.

7 0

2 years ago

5 points for the whole problem (Sorry it's all I have left)

Strike441 [17]

Answer: i don't see the problem

Step-by-step explanation:

3 0

3 years ago

A 95% confidence interval was computed using a sample of 16 lithium batteries, which had a sample mean life of 645 hours. The co

polet [3.4K]

Answer:

Step-by-step explanation:

Hello!

The mean life of 16 lithium batteries was estimated with a 95% CI:

(628.5, 661.5) hours

Assuming that the variable "X: Duration time (life) of a lithium battery(hours)" has a normal distribution and the statistic used to estimate the population mean was s Student's t, the formula for the interval is:

[X[bar]± $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ]

The amplitude of the interval is calculated as:

a= Upper bond - Lower bond

a= [X[bar]+ $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ] -[X[bar]- $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ]

and the semiamplitude (d) is half the amplitude

d=(Upper bond - Lower bond)/2

d=([X[bar]+ $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ] -[X[bar]- $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$ ] )/2

d= $t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }$

The sample mean marks where the center of the calculated interval will be. The terms of the formula that affect the width or amplitude of the interval is the value of the statistic, the sample standard deviation and the sample size.

Using the semiamplitude of the interval I'll analyze each one of the posibilities to see wich one will result in an increase of its amplitude.

Original interval:

Amplitude: a= 661.5 - 628.5= 33

semiamplitude d=a/2= 33/2= 16.5

1) Having a sample with a larger standard deviation.

The standard deviation has a direct relationship with the semiamplitude of the interval, if you increase the standard deviation, it will increase the semiamplitude of the CI

↑d= $t_{n-1;1-\alpha /2}$ * ↑S/√n

2) Using a 99% confidence level instead of 95%.

d= $t_{n_1;1-\alpha /2}$ * S/√n

Increasing the confidence level increases the value of t you will use for the interval and therefore increases the semiamplitude:

95% ⇒ $t_{15;0.975}= 2.131$

99% ⇒ $t_{15;0.995}= 2.947$

The confidence level and the semiamplitude have a direct relationship:

↑d= ↑ $t_{n_1;1-\alpha /2}$ * S/√n

3) Removing an outlier from the data.

Removing one outlier has two different effects:

1) the sample size is reduced in one (from 16 batteries to 15 batteries)

2) especially if the outlier is far away from the rest of the sample, the standard deviation will decrease when you take it out.

In this particular case, the modification of the standard deviation will have a higher impact in the semiamplitude of the interval than the modification of the sample size (just one unit change is negligible)

↓d= $t_{n_1;1-\alpha /2}$ * ↓S/√n

Since the standard deviation and the semiamplitude have a direct relationship, decreasing S will cause d to decrease.

4) Using a 90% confidence level instead of 95%.

↓d= ↓ $t_{n_1;1-\alpha /2}$ * S/√n

Using a lower confidence level will decrease the value of t used to calculate the interval and thus decrease the semiamplitude.

5) Testing 10 batteries instead of 16. and 6) Testing 24 batteries instead of 16.

The sample size has an indirect relationship with the semiamplitude if the interval, meaning that if you increase n, the semiamplitude will decrease but if you decrease n then the semiamplitude will increase:

From 16 batteries to 10 batteries: ↑d= $t_{n_1;1-\alpha /2}$ * S/√↓n

From 16 batteries to 24 batteries: ↓d= $t_{n_1;1-\alpha /2}$ * S/√↑n

I hope this helps!

4 0

3 years ago