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1 year ago

AB is a straight line workout the size of angle X

Mathematics

1 answer:

Hoochie [10]1 year ago

4 0

The x value will be 109°. The straight line formed a 180° angle. Solving the equation yields the angle.

<h3>What are supplementary angles?</h3>

Supplementary angles are two angels whose sum is 180°. When a straight line intersects a line, two angles form on each of the sides of the considered straight line.

Those two-two angles are supplementary angles in two pairs. That is, if two supplementary angles are adjacent to each other, their exterior sides form a straight line.

The straight line formed a 180° angle. The resulting equation is as follows:

⇒x+42°+29°=180°

⇒x=109°

Hence, the value of the x will be 109°

The complete question is:

AB is a straight line.

Work out the size of angle x.

Not drawn accurately

42°

29°

To learn more about supplementary angles, refer to:

brainly.com/question/12919120

#SPJ1

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If each youth saves $10 daily, their weekly savings would be $1400. So they donated $1400 to the church.

Step-by-step explanation:

Let the daily pay plan for each youth be $y

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Daily savings for 20 youth = 20×$10= $200

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3 years ago

A rabbit population doubles every 4 weeks. There are currently five rabbits in a restricted area. If t represents the time, in w

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3 years ago

The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de

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Answer:

a) $P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b) $P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c) $P(A') = 1-P(A) =1-0.4=0.6$

d) $P(A \cup B) =0.4 +0.8-0.2 =1.0$

e) The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

Step-by-step explanation:

We have the following space provided:

$S= [a,b,c,d,e]$

With the following probabilities:

$P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2$

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

$P(A) = 0.1+0.1+0.2 = 0.4$

$P(B) =0.2+0.4+0.2=0.8$

Determine:

a. P(A)

$P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b. P(B)

$P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c. P(A’)

From definition of complement we have this:

$P(A') = 1-P(A) =1-0.4=0.6$

d. P(AUB)

Using the total law of probability we got:

$P(A \cup B) =P(A) +P(B)-P(A \cap B)$

For this case $P(A \cap B) = P(c) =0.2$ , so if we replace we got:

$P(A \cup B) =0.4 +0.8-0.2 =1.0$

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

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3 years ago

If n(A) = P and n(B)=q then n(AxB) is (a ) p (b ) q ( c ) p+q ( d ) pq

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By way of example, suppose A = {1, 2, 3} and B = {a, b, c}. Then the Cartesian product of A and B is

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That is, each element in A gets a pairing with each element in B, and for each pairing you have n(A) choices for the first element and n(B) choices for the second element.

So if n(A) = p and n(B) = q, then n(A × B) = pq.

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3 years ago