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3 years ago

What is the area of the following parallelogram in square centimeters?

Mathematics

2 answers:

asambeis [7]3 years ago

7 0

Answer: For the firstone i think it is A but idk for the second one.

Step-by-step explanation:

balu736 [363]3 years ago

4 0

Answer with explanation:

1.⇒Area of Parallelogram = Corresponding height × Base

= 2 cm × 18 mm

→1 cm = 10 mm

So, Area of Parallelogram

= 2 cm × 1.8 cm

= 3.6 cm²

Option C: →3.6 cm²

→→→2. A triplet ,that is set of three numbers is called Pythagorean triplet, if it is of the form of : {m²-1, 2 m , m²+1}

That is if you write the Pythagorean rule for Right triangle, having sides , m²-1, 2 m , m²+1

then ,→ Square of Largest side = Sum of squares of two smallest side

Square of Largest Side = (m²+1)²

Sum of squares of two smallest side= (2 m)² + (m²-1)²

= 4 m² + (m²)² -2 m²+1

= (m²)² +2 m²+1

=(m²+1)²

So, we will check out , out of four options which of them are Pythagorean Triplet.

Option 1.

Square of Largest Number = 82²=6724

Sum of squares of two smaller number= 42² + 40²

= 1764 +1600

= 3364

→Square of Largest Number ≠Sum of squares of two smaller number

6724 ≠ 3364

Not a Pythagorean Triple.

Option 2.

Square of Largest Number = 58²=3364

Sum of squares of two smaller number= 42² + 40²

= 1764 +1600

= 3364

→Square of Largest Number =Sum of squares of two smaller number

3364= 3364

A Pythagorean Triple.

Option 3.

Square of Largest Number = 58²=3364

Sum of squares of two smaller number= 41² + 41²

= 1681 +1681

= 3362

→Square of Largest Number ≠Sum of squares of two smaller number

3364 ≠ 3362

Not a Pythagorean Triple.

Option 4.

Square of Largest Number = 57²=3249

Sum of squares of two smaller number= 41² + 40²

= 1681 +1600

= 3281

→Square of Largest Number ≠Sum of squares of two smaller number

3249 ≠ 3281

Not a Pythagorean Triple.

Option B: (42, 40,58) is a Pythagorean triple.

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$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$ In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$ And the variance is given by:

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Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

For this case we have the following distribution given:

X 3 4 5 6

P(X) 0.07 0.4 0.25 0.28

We can calculate the mean with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$

In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$

And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be: