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ValentinkaMS [17]
2 years ago
11

IXL Question! Brainliest + 5-star!

Mathematics
1 answer:
Nitella [24]2 years ago
4 0

Step-by-step explanation:

First, count the total number of the boxes from the figure. This can be done easily by counting the number of the rows and then the number of the columns and multiplying the numbers.

=> 10 x 10 = 100 boxes.

Now, count the shaded boxes. It's 8.

Now, we calculate percentage by,

=> Selected boxes/Total boxes x 100

Putting the values,

=> 8/100 x 100

=> 8%

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>:</em><em>)</em>

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Solve x/12+3&lt;7<br> thanks!
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Answer:

x < 48

Step-by-step explanation:

First, subtract 3 from both sides of this inequality:  x/12 < 4.

Multiply both sides of this inequality by 12 to remove fractions:  x < 48


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Solve the formula d = rt for t. A. t = rd B. t = d – r C. t = D. t =
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Answer:

(-4,0)

Step-by-step explanation:

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We go up 0.  That is the y coordinate

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4 years ago
After computing a confidence interval, the user believes the results are meaningless because the width of the interval is too la
navik [9.2K]

Answer:

The best recommendation is to: Increase the sample size.

Step-by-step explanation:

The width of a confidence interval  W=2\times CV\times \frac{SD}{\sqrt{n}} ,depends on three things,

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  2. Standard deviation
  3. Confidence level

To decrease the width the following options can be used:

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Since the sample is inversely proportional to the width of the confidence interval, increasing the value of <em>n</em> will decrease the width.

  • Decrease the standard deviation.

The standard deviation is directly proportional to the width of the confidence interval. On decreasing the standard deviation value the width of the interval will also decrease.

  • Decrease the confidence level.

The critical value of the distribution is based on the confidence level. Higher the confidence level, higher will be critical value.

So, on deceasing the confidence level the critical value will decrease, hence decreasing the width of the interval.

Thus, the best recommendation is to: Increase the sample size.

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3 years ago
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