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2 years ago

Only answer if you know for sure you’re right! :)

Mathematics

2 answers:

shutvik [7]2 years ago

8 0

2(7x+7)...................

algol [13]2 years ago

6 0

Answer:

14x+14=14(x+1)

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What is the mean absolute deviation of the data set? Round to the nearest hundredth if necessary.

hammer [34]

The mean absolute deviation, rounded to the nearest hundredth, is: 11.67 calories.

<h3>How to Find the Mean Absolute Deviation?</h3>

To find the mean absolute deviation, find the mean of the data then find the average distance of each data point from the mean.

Given the data, 61, 42, 52, 27, 35, 23:

Mean = (61 + 42 + 52 + 27 + 35 + 23)/6

Mean = 240/6

Mean = 40

Mean absolute deviation = [(61 - 40) + ( 42 - 40) + ( 52 - 40) + ( 27 - 40) + ( 35 - 40) + ( 23 - 40) 6 = (21) + (2) + (12) + (13) + (5) + (17)] / 6 = 70/6

Mean Absolute Deviation = 11.67 calories.

Learn more about mean absolute deviation on:

brainly.com/question/447169

#SPJ1

6 0

1 year ago

A family ordered some candy bars and bags of chips for snacking. The number of candy bars was twice as much as the number of bag

sammy [17]

Answer:

The family bought approximately 2 bags of chips.

Step-by-step explanation:

Given:

Total spent by family = $5.50

We need to find bags of chips did the family buy.

Solution:

Let number of bags of chips bought be 'x'.

Also Given:

The number of candy bars was twice as much as the number of bags of chips.

number of candy bars = $2x$

Now we can say that;

Total spent by family is equal to sum of the number of bags of chips bought and number of candy bars.

framing in equation form we get;

$x+2x=5.5\\\\3x=5.5$

Dividing both side by 3 we get;

$\frac{3x}{3}=\frac{5.5}{3}\\\\x=1.83\approx2$

Hence the family bought approximately 2 bags of chips.

7 0

3 years ago

Assuming that the population is normally distributed, construct a 9090% confidence interval for the population mean for each o

jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

Set B:

Mean: 4.5

Standard dev: 2.45

% = 90

Set A:

Standard Error, SE = s/ √n = 1.9/√8 = 0.67

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

t- score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

Set B:

Standard Error, SE = s/ √n = 2.45/√8 = 0.87

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

t- Score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.

6 0

3 years ago

4/5 of a number is 16. What is the number?

Zepler [3.9K]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Shirley graphed the line shown on the coordinate plane

Rus_ich [418]

But this is not a question. proof? no question mark

8 0

2 years ago