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balu736 [363]
3 years ago
7

Figure ABCD is a trapezoid find the Value of x

Mathematics
2 answers:
denpristay [2]3 years ago
4 0

Answer:

<em>The value of x is 4.</em>

Step-by-step explanation:

Figure ABCD is a trapezoid, in which AD and BC are parallel sides.

Given that, AD= 3x+10 and BC=2x

Now, <u>the length of the median</u>  = \frac{1}{2}(AD+BC) , <u>which is given as</u>  15.

So, the equation will be......

\frac{1}{2}[(3x+10)+2x]=15\\ \\ \frac{1}{2}(5x+10)=15\\ \\ 5x+10=2(15)\\ \\ 5x+10=30\\ \\ 5x=30-10\\ \\ 5x=20\\ \\ x=\frac{20}{5}=4

So, the value of x is 4.

valina [46]3 years ago
3 0

Answer:

The value of x is 4

Step-by-step explanation:

we know that

in this problem

\frac{1}{2}(2x+(3x+10))=15

solve for x

\frac{1}{2}(5x+10)=15

(5x+10)=30

5x=30-10

x=20/5=4

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son4ous [18]

Answer:

14 :)

Step-by-step explanation:

all you need to do is add the ones

3 0
2 years ago
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Write the elements belonging to the set {x|x is a natural number less than 10
Feliz [49]

Answer:

The set A = { 1,2,3,4,5,6,7,8,9}

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the set

 { x/x/ is a natural number less than 10 }

<u><em>Natural numbers :</em></u>

The natural numbers are

          { 1,2,3,4,5,6,7,8,9,10,.........}

The set A =  { x/x/ is a natural number less than 10 }

The set A = { 1,2,3,4,5,6,7,8,9}

4 0
3 years ago
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp
olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = \overline{x} = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data​ speeds i.e. x - \overline{x}

x - \overline{x} = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

\frac{59.98}{20.03}=2.99

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

z=\frac{x-\overline{x}}{s}

Using the given values, we get:

z=\frac{75.6-15.62}{20.03}=2.99

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly​ high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the  carrier's highest data speed​ is significantly higher.

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15x18 =area
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5 (x + 7) is the answer
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