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Anettt [7]
3 years ago
8

The dot plot shows the number of flowers each science student was able to identify. A dot plot whose number line is labeled Numb

er of flowers identified. There are 2 dots above 0, 4 dots above 1, 5 dots above 2, 7 dots above 3, 4 dots above 4, 3 dots above 5, 2 dots above 6, 1 dot above 7, 2 dots above 8, 1 dot above 9, 1 dot above 10, and 1 dot above 12. Select from the drop-down menus to correctly complete the statement. The distribution of data is skewed right , with a mode of Choose... , and a range of Choose... .
Mathematics
1 answer:
irina [24]3 years ago
3 0

Answer:

The dot plot shows the number of flowers each science student was able to identify.

A dot plot whose number line is labeled Number of flowers identified. There are 2 dots above 0, 4 dots above 1, 5 dots above 2, 7 dots above 3, 4 dots above 4, 3 dots above 5, 2 dots above 6, 1 dot above 7, 2 dots above 8, 1 dot above 9, 1 dot above 10, and 1 dot above 12.

Select from the drop-down menus to correctly complete the statement.

The distribution of data is 1 "skewed right", with a mode of 2 "3", and a range of 3 "12".

Step-by-step explanation:

I took the test!

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Use the like bases property to solve the equation:<br> 16^x = 1/4<br> x=
gavmur [86]

Answer:-1/2

Step-by-step explanation:

16^x=1/4

(4^2)^x=4^(-1)

4^2x=4^(-1)

Base cancelled each other since they are the same

2x=-1

Divide both sides by 2

2x/2=-1/2

x=-1/2

7 0
3 years ago
What is the surface area of the figure?
Whitepunk [10]
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How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:  

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}

Where n = 8 and r = 0,1....8

When r = 0

\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1

When r = 1

\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 2

\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8

When r = 3

\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 4

\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70

When r = 5

\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 8

\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

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3 years ago
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Wittaler [7]
<h3>Two Answers: choice C, choice D</h3>

Look at where we don't have repeating x values. This happens with function C and function D. All the x values are unique for each choice mentioned.

In choices A, B, and E, the value x = -3 repeats itself. So we don't have a function for either of these. A function is only possible if any input (x) leads to exactly one output (y).

6 0
3 years ago
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