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ad-work [718]
3 years ago
5

What type of polygon is pictured here

Mathematics
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

irregular heptagon

Step-by-step explanation:

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PLEASE EASY HELP*****
frez [133]

Answer:

B

Step-by-step explanation:

Multiply through by x+5

2(x + 1) = x + 5 - 1            Combine like terms

2x + 2 = x + 4                 Subtract 2 from both sides.

2x = x + 4 - 2

2x = x + 2                         Subtract x from both sides.

2x - x = 2                          Combine

x = 2

B

6 0
3 years ago
Read 2 more answers
A car costs$cents when new. It was sold for four fifths of its cost price. How much money was lost on the car.
REY [17]

The question is incomplete as the cost price isn't given. However, taking the cost price as x :

Answer:

Kindly check explanation

Step-by-step explanation:

Given :

A car costs$cents when new. It was sold for four fifths of its cost price. How much money was lost on the car.

Let :

Cost price when new = x

Cost price when sold = 4/5 * cost price when new

Cost when sold = 4/5 of x = 4x/5

Amount of money lost on the car = (Cost price of car when new - Cost of car when sold)

Hence,

Amount of money lost on the car = (x - 4x/5)

x - 4x/5 = (5x - 4x) / 5 = x / 5

To obtain the exact price, kindly input the omitted cost when new for x.

7 0
2 years ago
A statistics tutor wants to assess whether her remedial tutoring has been effective for her five students. She decides to conduc
Snezhnost [94]

Answer:

The test statistic value is, <em>t</em> = -5.245.

The effect size using estimated Cohen's d is 2.35.

Step-by-step explanation:

A paired <em>t</em>-test would be used to determine whether the remedial tutoring has been effective for the statistics tutor's five students.

The hypothesis can be defined as follows:

<em>H</em>₀: The remedial tutoring has not been effective, i.e. <em>d</em> = 0.

<em>Hₐ</em>: The remedial tutoring has been effective, i.e. <em>d</em> > 0.

Use Excel to perform the Paired <em>t</em> test.

Go to Data → Data Analysis → t-test: Paired Two Sample Means

A dialog box will open.

Select the values of the variables accordingly.

The Excel output is attached below.

The test statistic value is, <em>t</em> = -5.245.

Compute the effect size using estimated Cohen's d as follows:

\text{Cohen's d}=\frac{Mean_{d}}{SD_{d}}

                =\frac{0.54}{0.23022}\\\\=2.34558\\\\\approx 2.35

Thus, the effect size using estimated Cohen's d is 2.35.

8 0
3 years ago
Answer this question please I need help
faust18 [17]
14872 is the answer.
4 0
3 years ago
First photo question-The graph below shows three different normal distributions.
zheka24 [161]

Answer:

For the <u>first question</u>:  They must each have a different mean and a different standard deviation.  For the <u>second question</u>:  In an observational study the researchers do not control treatment and in an experiment they do.  For the <u>third question</u>:  The second graph.  For the <u>fourth question</u>:  The second graph.

Step-by-step explanation:

For the <u>first question</u>:  We can see that each bell-shaped curve is centered around a different number.  This tells us that their means are all different.

We can also see that each curve has a different width; this means that the spread of data, measured in part by the standard deviation, is different as well.  This means both the mean and the standard deviation of each distribution is different.

For the <u>second question</u>:  In an observational study, as the name implies, all that is done is observation.  There is no treatment given by the researchers.  In an experiment, however, one group receives treatment and one group does not, so the differences can be measured.

For the <u>third question</u>:  In the first graph, the probabilities add up to 1.2.  This is more than 100%, so this is not a valid distribution.  In the second graph, the probabilities add up to 1.0; this is 100% exactly, so this is a valid distribution.

For the <u>fourth question</u>:  In the first graph, the probabilities only add up to 0.9.  This is not 100%, so this is not a valid distribution.  In the second graph, the probabilities add up to 1.0; this is 100% exactly, so this is a valid distribution.

8 0
3 years ago
Read 2 more answers
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