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Veseljchak [2.6K]

4 years ago

12

zahra has paper rectangles of different sizes. every rectangle is 5 cm longer than it is wide. Is there a proportional relations

hip between the lengths and widths of these rectangles? Explain.

2 answers:

Anastaziya [24]4 years ago

6 0

Answer:

No, there is no proportional relationship exist between the lengths and widths of these rectangles.

Step-by-step explanation:

Since, it is given that every rectangle is 5 cm longer than it's width.

⇒ If Width = W, then Length = W + 5

Also, she has a rectangle of different sizes.

Thus, there can't be more relationship exist between Length and Width.

The only relationship between Length and Width is:

W = L + 5

where, W is Width of rectangle.

and, L is Length of rectangle.

Vinil7 [7]4 years ago

3 0

As given Zahra has paper rectangles of different sizes.

Also, Every rectangle is 5 cm is longer than it's breadth.

So, if Length=L, then Breadth= L- 5

Or , if Breadth= B, then Length= B+5

or, if length is x and breadth is y ,then writing in terms of equation

→x=y+5

As you can see,there is not any proportional relationship between length and widths of these rectangles.

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A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws. Use a histogram of the binomial dis

DiKsa [7]

Answer:

The most likely outcome is exactly 4 free throws

Step-by-step explanation:

Given

$n = 5$ --- attempts

$p = 82.6\%$ ---- probability of a successful free throw

$p = 0.826$

Required

A histogram to show the most likely outcome

From the question, we understand that the distribution is binomial.

This is represented as:

$P(X = x) = ^nC_x * p^x * (1 - p)^{n-x}$

For x = 0 to 5, where x represents the number of free throws; we have:

$P(X = x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(X = 0) = ^5C_0 * 0.826^0 * (1 - 0.826)^{5-0}$

$P(X = 0) = ^5C_0 * 0.826^0 * (0.174)^{5}$

$P(X = 0) = 1 * 1 * 0.000159 \approx 0.0002$

$P(X = 1) = ^5C_1 * 0.826^1 * (1 - 0.826)^{5-1}$

$P(X = 1) = ^5C_1 * 0.826^1 * (0.174)^4$

$P(X = 1) = 5 * 0.826 * 0.000917 \approx 0.0038$

$P(X = 2) = ^5C_2 * 0.826^2 * (1 - 0.826)^{5-2}$

$P(X = 2) = ^5C_2 * 0.826^2 * (0.174)^{3}$

$P(X = 2) = 10 * 0.682 * 0.005268 \approx 0.0359$

$P(X = 3) = ^5C_3 * 0.826^3 * (1 - 0.826)^{5-3}$

$P(X = 3) = ^5C_3 * 0.826^3 * (0.174)^2$

$P(X = 3) = 10 * 0.5636 * 0.030276 \approx 0.1706$

$P(X = 4) = ^5C_4 * 0.826^4 * (1 - 0.826)^{5-4}$

$P(X = 4) = 5 * 0.826^4 * (0.174)^1$

$P(X = 4) = 5 * 0.4655 * 0.174 \approx 0.4050$

$P(X = 5) = ^5C_5 * 0.826^5 * (1 - 0.826)^{5-5}\\$

$P(X = 5) = ^5C_5 * 0.826^5 * (0.174)^0$

$P(X = 5) = 1 * 0.3845 * 1 \approx 0.3845$

From the above computations, we have:

$P(X = 0) \approx 0.0002$

$P(X = 1) \approx 0.0038$

$P(X = 2) \approx 0.0359$

$P(X = 3) \approx 0.1706$

$P(X = 4) \approx 0.4050$

$P(X = 5) \approx 0.3845$

See attachment for histogram

<em>From the histogram, we can see that the most likely outcome is at: x = 4</em>

<em>Because it has the longest vertical bar (0.4050 or 40.5%)</em>

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