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3 years ago

Can the segments of 6 5 and 12 form a triangle

Mathematics

1 answer:

Nadya [2.5K]3 years ago

5 0

They can form the Triangle

But, they can't form right angled triangle, because in that case, largest side should be smaller than the sum of other two........

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WELP ME PLZZ THANKSSS

Luden [163]

Answer:

I got 32169.91

Step-by-step explanation:

so I would put D bc it's closest

Hope This Helps! Have A Nice Day!!

5 0

3 years ago

How do I make both these equations into a elimination equation? Look at the pic I’m very confused!

qwelly [4]

Answer:

Step-by-step explanation:

c + a = 342 ..................... (1)

5c + 12a = 2550 ........... (2)

- 5 × (1) + (2)

7a = 840

a = 120

c = 342 - 120 = 222

6 0

3 years ago

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

3 years ago

6x = 2y - 7 what is the terms

Fudgin [204]

Answer:

y = 3x+7/2, and x = (2y-7)/6

Step-by-step explanation:

6 0

3 years ago

!!HURRY IM BEING TIMED 40PTS!!

Rudik [331]

The average rate of change in a function from a to b is

$\frac{f(b)-f(a)}{b-a}$ .

For the left function, from x = 4 to x = 5, the rate of change is

$\frac{100-64}{5-4} = \frac{36}{1} = 36$ .

For the right function, the rate of change is

$\frac{1024-256}{5-4} = \frac{768}{1} = 768$

Using a ratio to compare the right function's growth rate with that of the left function, you get $768 \div 36 \approx 21.3$ .

The right function is growing approximately 21 times faster than the left function (over the interval from 4 to 5).

3 0

3 years ago

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