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3 years ago

Please help! Due Friday

Mathematics

2 answers:

solmaris [256]3 years ago

5 0

For (-1,2) you go to the left one and up two.
For (-3,6) left three up six
For (4,8) right four and up eight

KiRa [710]3 years ago

3 0

Answer:

So for each point's numbers you have the x and the y. The first is the x, 2nd y in the parenthesis.

On the graph you start with the x axis, x in the middle is 0 to the left is the negative #'s and on the right: positive. Y axis is top positive, bottom negative! Now put the numbers on the axis' and then plot the points!

Hope this helps!!! :)

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A public health organization reports that 40%of baby boys 6-8 months old in the United

Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

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.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then:

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

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502$

Hence:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675$

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106$

$P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016$

$P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001$

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

4 0

2 years ago

Nathan deposited 7/9 of his allowance into his savings account. He spent the remaining amount, or $2.50. How much did Nathan dep

inysia [295]

2/9=2.50
Allowence=2.50x9 divided by which gives you an answer of 11.25
Hope this helps you!

4 0

3 years ago

I really need your help . And don’t worry this is just a practice :)

MariettaO [177]

Answer:

65/80 score

Step-by-step explanation:

3 0

3 years ago

Refer to the sample data for pre-employment drug screening shown below. If one of the subjects is randomly selected, what is t

RoseWind [281]

Answer:

0.193877

Step-by-step explanation:

The data given to us is

<em><u>Pre-Employment Drug Screening Results</u></em>

Positive test result Negative test result

Drug Use Is Indicated Drug Use Is Not Indicated

Subject Uses Drugs: 38 12

Subject Is Not a drug user: 19 29

Now the total of this is = 38+19+12+29= 98

Now the probability of false positive is = 19/98= 0.193877

The <u>Subject Is Not a drug user </u> would suffer from a false positive. He is not a user and has a positive result.

7 0

3 years ago

Please Help I will reward brainliest

madreJ [45]

1/
V=44.312 in^3
2/
V=42.453 in^3
3/
V=75.36 in^3
4/
V=696.557 in^3
5/
V=671.175 in^3. Hope it help!

8 0

3 years ago