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2 years ago

Pls help!!

Mathematics

1 answer:

sleet_krkn [62]2 years ago

3 0

Answer:

See below

Step-by-step explanation:

The CENTER will de directly between the two given points

so...so far the circle eq is

(x+1)^2 + ( y-7)^2

the RADIUS is 1/2 of the distance between the given points

1/2 d = (using distance formula) = 1/2 sqrt 80 =<u> 4.472 </u>

then the equation of the circle in standard form

(x+1)^2 + (y-7)^2 = 20

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Mary checked a book out of the public library two days ago, and she read half of the book that night. yesterday, she read anothe

ehidna [41]

Answer:

Since she read half of the book, it's 50%+ the 40% she read yesterday, that make 90%.So she is left with 10%

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2 years ago

The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.

mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

0.96516 probability of a chipmunk living through the year, thus $p = 0.96516$

Item a:

Two is P(X = 2) when n = 2, thus:

$P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315$

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

Six is P(X = 6) when n = 6, then:

$P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834$

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

At least one not living is:

$P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166$

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

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3 years ago

Solve x^2-y^2+x+1+2yx+y=0

Harlamova29_29 [7]

Combine like terms and use inverse operations

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3 years ago

Find the indicated limit, if it exists. limit of f of x as x approaches 0 where f of x equals 5x minus 8 when x is less than 0 a

vlabodo [156]

The limit does not exist. Why? Because the left hand limit DOES NOT equal the right hand limit. Let’s double check:

We could use -0.000001 to represent the left hand limit. This is less than 0. We plug in 5x - 8

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-0.000005 - 8
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If we would continue the limit (extend the zeros to infinity), we would get exactly
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That is our left hand limit.

Our right hand limit will be represented by 0.000001. This is greater than 0. We plug in abs(-4 - x)

abs(-4 - (0.000001))
abs(-4.000001)
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If we would continue the limit (extend the zeros to infinity), we would get exactly
4

4 does not equal -8, therefore

The limit does not exist

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Answer:4

Step-by-step explanation:

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