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3 years ago

Help pleaseeeee:)))))))

Mathematics

2 answers:

omeli [17]3 years ago

8 0

5 by 3

3rd one

there we have to find hypotenuse.

Inga [223]3 years ago

3 0

Answer:

+- 5/3

Step-by-step explanation:

When we construct a right triangle with legs with length 3 and 4. The hypoteneuse must 5. Since tan(x) = opp/adj = 4/3 and sec(x) = hyp/adj,

We get the answer to be 5/3. Now there is a +- there because if we use the unit circle, if we take the point (-3, -4), we get tan(x) = (-4/-3) = 4/3. However, the hypoteneuse for sec(x) is always positive so sec(x) = 5/-3 = -5/3.

That is why there are two solutions, one with the terminal arm in Q3 and one in Q1. But that's the explanation using the unit circle.

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The probability that Barry Bonds hits a home run on any given at-bat is 0.16, and each at-bat is independent.

serious [3.7K]

Answer:

a) 0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.

b) The expected number of at-bats until the next home run is 6.25.

Step-by-step explanation:

For each at bat, there are two possible outcomes. Either it is a home run, or it is not. The probability of an at bat resulting in a home run is independent of any other at-bat, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that Barry Bonds hits a home run on any given at-bat is 0.16

This means that $p = 0.16$

Part A: What is the probability that the next home run will be on his fifth at-bat?

0 on his next 4(P(X = 0) when n = 4)

Home run on his 5th at-bat, with 0.16 probability. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.16)^{0}.(0.84)^{4} = 0.49787136
$

0.49787136 *0.16 = 0.0797.

0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.

Part B: What is the expected number of at-bats until the next home run?

The expected number of trials for n successes is given by:

$E = \frac{n}{p}$

In this question, $n = 1, p = 0.16$ . So

$E = \frac{1}{0.16} = 6.25$

The expected number of at-bats until the next home run is 6.25.

6 0

2 years ago

Two distinct planes __________ intersect in a pair of lines

hammer [34]

If the planes are parallel to one another then they will never intersect (assuming they are distinct planes as the question states).

Otherwise they will intersect along a single line.

Therefore the answer is NEVER - they can never intersect in a pair of lines.

3 0

3 years ago

7th grade math PLEASE HELP FOR BRAINLIEST ASAP ASAP ASAP PLEASEEEEEEEEEEEEE I NEEED THE HELPPPP, :):):):):):):):):):):)

LUCKY_DIMON [66]

Answer:

area=3.14*2.3*2.3

=3.14*5.29

=16.6106

6 0

3 years ago

Read 2 more answers

Helpppp plssss!!!<br><br>Appreciate you!!