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3 years ago

I NEEEEED HELP AND QUICK T~T

Mathematics

1 answer:

dsp733 years ago

8 0

Answer:

5. Alt ext, 6. Alt int, 7. Alt ext, 8, SS Int, 9. Corr

Step-by-step explanation:

5. Alternate exterior since they are on opposite sides of the transversal but both are exterior to parallel lines.

6. Alternate interior, opposite sides of transversal but both are inside the parallel lines.

7. Alternate exterior, same as #1

8. Same side interior since they are on the same side of the transversal and both are inside the parallel lines

9. Corresponding since they are at the same place on the 2 parallel lines.

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Assume that the probability that a baseball player will get a hit in any one at bat is 0.250. What is the probability that he wi

Yanka [14]

Answer: 0.0791

Step-by-step explanation:

Given : The probability that a baseball player will get a hit in any one at bat is 0.250.

Let x be the first hit.

According to the geometric probability , the probability that x (th) trial is the first success = $P(X=x)=(1-p)^{x-1}p$ , where p is the probability of getting success in each trial.

As per given p= 0.250

Then, the probability that he will get his first hit on his 5th at bat will be :

$P(X=x)=(1-0.25)^{5-1}(0.25)\\\\=(0.75)^4(0.25)\\\\=0.0791015625\approx0.0791$

Hence, the probability that he will get his first hit on his 5th at bat is 0.0791 .

7 0

3 years ago

Write a real world problem that can be represented by the equation y+20=85

madam [21]

Hanna bought 85 candies and ate 20 of them. How many candies are left?

5 0

3 years ago

Read 2 more answers

Please include work

OLga [1]

Answer:

9. 6k + 5 and 2k - 1.

10. 5√2 .

Step-by-step explanation:

9. Area = 12k^2 + 4k - 5

Factoring:

12k^2 - 6k + 10k - 5

= 6k(2k - 1) + 5(2k - 1)

= (6k + 5)(2k - 1)

So the required lengths are 6k + 5 and 2k - 1.

10. The ratio of a leg to the hypotenuse is 1: √2 so if the hypotenus = 10 the length of a leg is 10 / √2

= 10 √2 / 2

= 5√2 .

7 0

4 years ago

Simplify completely..........

vitfil [10]

Answer:

$\frac{x}{x-1}$

Step-by-step explanation:

$\frac{3x^2 - 1}{x^2 - 1} - \frac{2x + 1}{x + 1}$ $[ \ x^ 2- 1 = (x-1)(x + 1) \ ]$

$= \frac{3x^2 - 1}{(x-1)(x + 1 )} - \frac{2x + 1}{x + 1}\\\\=\frac{(3x^2 - 1)-(2x + 1)(x-1)}{(x + 1)(x-1)}$ $[\ Taking \ LCM \ ]$

$= \frac{(3x^2 - 1)- (2x^2 - 2x + x - 1)}{(x+1)(x-1)}\\\\=\frac{3x^2 - 1 - 2x^2 + x + 1 }{(x+1)(x-1)}\\\\=\frac{x^2 +x}{(x+1)(x-1)}\\\\=\frac{x(x+1)}{(x+1)(x-1)}\\\\=\frac{x}{x-1}$