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Nesterboy [21]
3 years ago
8

ABDC is a trapezoid. Line AB = 2.5, and Line EF = 4.5. What is the length of Line CD ?

Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0
The line CD is the median of thre trapezoid, therefore its measures is shown below:
 You must sum the largest base and the shortest base and divide both of them by 2, as following:
 CD=(4.5+2.5)/2
 CD=7/2
 CD=3.5
 Therefore, as you can see, the measure of the line CD is 3.5
 The answer is: 3.5
mezya [45]3 years ago
5 0
The correct answer is 3.5

A trapezoid is a 4-sided flat shape with straight sides. This shape has a pair of opposite sides parallel. So according to the figure we know the following:

\overline{AC}=\overline{CE}

And it is also true that:

\overline{BD}=\overline{EF}

We know that the lengths of the parallel sides are given by:

\overline{AB}=2.5 \\ \\ \overline{EF}=4.5

So the goal is to find the length of Line CD. This line is in fact the <em>median</em> of the trapezoid <span>(also called a midline or midsegment) that is a line segment half-way between the two bases and it can be found as follows:
</span>
m=\frac{\overline{AB}+\overline{EF}}{2}=\frac{2.5+4.5}{2}=\frac{7}{2} \\ \\ \therefore \boxed{\overline{CD}=m=3.5}
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allochka39001 [22]

Answer:

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Step-by-step explanation:

Given

Point 1 (1,2)

Point 2 (5,8)

Required

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Though the graph would have assisted in answering the question; its unavailability doesn't mean the question cannot be solved.

Having said that,

the constant variation can be solved by calculating the gradient of the graph;

The gradient is often represented by m and is calculated as thus

m = \frac{y_2 - y_1}{x_2 - x_1}

Where

(x_1, y_1) = (1,2)\\(x_2, y_2) = (5,8)

By substituting values for x1,x2,y1 and y2; the gradient becomes

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3 years ago
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Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if:a) a
lara31 [8.8K]

Answer:

A) 0.0009765625

B) 0.0060466176

C) 2.7756 x 10^(-17)

Step-by-step explanation:

A) This problem follows a binomial distribution. The number of successes among a fixed number of trials is; n = 10

If a 0 bit and 1 bit are equally likely, then the probability to select in 1 bit is; p = 1/2 = 0.5

Now the definition of binomial probability is given by;

P(K = x) = C(n, k)•p^(k)•(1 - p)^(n - k)

Now, we want the definition of this probability at k = 10.

Thus;

P(x = 10) = C(10,10)•0.5^(10)•(1 - 0.5)^(10 - 10)

P(x = 10) = 0.0009765625

B) here we are given that p = 0.6 while n remains 10 and k = 10

Thus;

P(x = 10) = C(10,10)•0.6^(10)•(1 - 0.6)^(10 - 10)

P(x=10) = 0.0060466176

C) we are given that;

P((x_i) = 1) = 1/(2^(i))

Where i = 1,2,3.....,n

Now, the probability for the different bits is independent, so we can use multiplication rule for independent events which gives;

P(x = 10) = P((x_1) = 1)•P((x_2) = 1)•P((x_3) = 1)••P((x_4) = 1)•P((x_5) = 1)•P((x_6) = 1)•P((x_7) = 1)•P((x_8) = 1)•P((x_9) = 1)•P((x_10) = 1)

This gives;

P(x = 10) = [1/(2^(1))]•[1/(2^(2))]•[1/(2^(3))]•[1/(2^(4))]....•[1/(2^(10))]

This gives;

P(x = 10) = [1/(2^(55))]

P(x = 10) = 2.7756 x 10^(-17)

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