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3 years ago

Estimate the perimeter of the figure to the nearest whole number.

Mathematics

1 answer:

Paha777 [63]3 years ago

4 0

Answer:

The perimeter (to the nearest integer) is 9.

Step-by-step explanation:

The upper half of this figure is a triangle with height 3 and base 6. If we divide this vertically we get two congruent triangles of height 3 and base 3. Using the Pythagorean Theorem we find the length of the diagonal of one of these small triangles: (diagonal)^2 = 3^2 + 3^2, or (diagonal)^2 = 2*3^2.

Therefore the diagonal length is (diagonal) = 3√2, and thus the total length of the uppermost two sides of this figure is 6√2.

The lower half of the figure has the shape of a trapezoid. Its base is 4. Both to the left and to the right of the vertical centerline of this trapezoid is a triangle of base 1 and height 3; we need to find the length of the diagonal of one such triangle. Using the Pythagorean Theorem, we get

(diagonal)^2 = 1^2 + 3^2, or 1 + 9, or 10. Thus, the length of each diagonal is √10, and so two diagonals comes to 2√10.

Then the perimeter consists of the sum 2√10 + 4 + 6√2.

which, when done on a calculator, comes to 9.48. We must round this off to the nearest whole number, obtaining the final result 9.

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Option C:

The value of m is 4.

Solution:

Given data:

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<u>To find the value of m:</u>

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Answer:

(a) P(A) = 0.34

(b) P(B) = 0.33

(c) P(C) = 0.33

(d) The Complement of event A = 1 - P(A) = 0.66

Step-by-step explanation:

We are given that in the long run, roommate A chooses rock 36% of the time, and roommate B chooses rock 22% of the time; roommate A selects paper 32% of the time, and roommate B selects paper 25% of the time; roommate A chooses scissors 32% of the time, and roommate B chooses scissors 53% of the time.

Let the probability that roommate A chooses rock = P( $R_A$ ) = 0.36

The probability that roommate A chooses paper = P( $P_A$ ) = 0.32

The probability that roommate A chooses scissors = P( $S_A$ ) = 0.32

The probability that roommate B chooses rock = P( $R_B$ ) = 0.22

The probability that roommate B chooses paper = P( $P_B$ ) = 0.25

The probability that roommate B chooses scissors = P( $S_B$ ) = 0.53

(a) Let A = event that roommate A wins the game and thus does not have to wash the dishes.

This will happen only when roommate A chooses rock and roommate B chooses scissors or roommate A chooses paper and roommate B chooses rock or roommate A chooses scissors and roommate B chooses paper.

So, P(A) = $P(R_A) \times P(S_B) + P(P_A) \times P(R_B) + P(S_A) \times P(P_B)$

= (0.36 $\times$ 0.53) + (0.32 $\times$ 0.22) + (0.32 $\times$ 0.25)

= 0.1908 + 0.0704 + 0.08

P(A) = 0.34

(b) Let B = event that roommate B wins the game and thus does not have to wash the dishes.

This will happen only when roommate B chooses rock and roommate A chooses scissors or roommate B chooses paper and roommate A chooses rock or roommate B chooses scissors and roommate A chooses paper.

So, P(B) = $P(R_B) \times P(S_A) + P(P_B) \times P(R_A) + P(S_B) \times P(P_A)$

= (0.22 $\times$ 0.32) + (0.25 $\times$ 0.36) + (0.53 $\times$ 0.32)

= 0.0704 + 0.09 + 0.1696

P(B) = 0.33

(c) Let C = event that the game ends in a tie.

This will happen only when roommate A chooses rock and roommate B also chooses rock or roommate A chooses paper and roommate B also chooses paper or roommate A chooses scissors and roommate B also chooses scissors.

So, P(C) = $P(R_A) \times P(R_B) + P(P_A) \times P(P_B) + P(S_A) \times P(S_B)$

= (0.36 $\times$ 0.22) + (0.32 $\times$ 0.25) + (0.32 $\times$ 0.53)

= 0.0792 + 0.08 + 0.1696

P(C) = 0.3288 ≈ 0.33

(d) The complement of event A = P(A') = 1 - P(A)

= 1 - 0.34 = 0.66.

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