List two sets of possible side lengths for two triangles, one set that forms a triangle and one set that does not.

1 answer:

6 0

In any triangle, the sum of lengths of any two sides must be greater than the third side.

(1) If we chose three side lengths to be 5 cm, 7 cm and 10 cm. Let's do the check:
5+ 7 = 12 > 10
5+10 = 15 > 7
7 + 10 = 17 > 5
The sum of any two sides is greater than the third, therefore, these three side lengths can form a triangle.

(2) If we chose three side lengths to be 3 cm, 4 cm and 8 cm.Let's do the check:
3 + 4 = 7 which is not greater than 8
The condition is not fulfilled, therefore, these three side lengths cannot form a triangle.

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3 years ago

A minor league baseball team plays 128 games a seanson. If the tam won 16 more than three times as many games as they lost how m

tiny-mole [99]

Answer: The team won 100 games and losses 28 games

Step-by-step explanation:

Let the number of games won denoted by x and the games lost is denoted by y .

Given : A minor league baseball team plays 128 games in a season.

Team won 16 more than three times as many games as they lost.

Then, the system of equations, we have

$x+y=128-------(1)\\\\x=3y+16------------(2)$

We can rewrite equation (2) as $x-3y=16----------------(3)$

Now, Eliminate equation (3) from (1), we get

$y-(-3y)=128-16\\\\\Rightarrow\ 4y=112\\\\\Rightarrow\ y=\dfrac{112}{4}=28$

Substitute the value of y in (2), we get

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Hence, the team won 100 games and losses 28 games.

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3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$