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3 years ago

Can someone solve this

Mathematics

1 answer:

Natalka [10]3 years ago

4 0

Option D:

ΔCAN ≅ ΔWNA by SAS congruence rule.

Solution:

Given data:

m∠CNA = m∠WAN and CN = WA

To prove that ΔCAN ≅ ΔWNA:

In ΔCAN and ΔWNA,

CN = WA (given side)

∠CNA = ∠WAN (given angle)

NA = NA (reflexive side)

Therefore, ΔCAN ≅ ΔWNA by SAS congruence rule.

Hence option D is the correct answer.

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Answer:

Part A) The probability is closer to 0, because is less than 50%

Part B) The probability is closer to 1, because is greater than 50%

Step-by-step explanation:

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.

The size of the sample space is the total number of possible outcomes

The event space is the number of outcomes in the event you are interested in.

Let

x------> size of the event space

y-----> size of the sample space

$P=\frac{x}{y}$

Part A) Is the probability of hitting the black circle inside the target closer to 0 or 1?

In this problem we have

x------> is the area of the black circle

y------> is the total area of the square target

Find the area of black circle

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$x=(3.14)(1.5)^{2} =7.065\ units^{2}$

Find the total area of the square target

$y=9^{2}=81\ units^{2}$

Find the probability

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Convert to percentage

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The probability is closer to 0, because is less than 50%

Part B) Is the probability of hitting the white portion of the target closer to 0 or 1?

In this problem we have

x------> is the area of the white portion of the target

y------> is the total area of the square target

Find the area of the white portion

The area is equal to the area of the square minus the area of the circle

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Find the probability

$P=\frac{73.935}{81}=0.9128$

Convert to percentage

$0.9128*100=91.28\%$

The probability is closer to 1, because is greater than 50%

4 0

3 years ago