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Artyom0805 [142]
3 years ago
7

DO THE MATH NOW

Mathematics
1 answer:
Bond [772]3 years ago
6 0

no because 40-0.05

<em>dssw</em><em> </em><em>its</em><em> </em><em>really</em><em> </em><em>assey</em>

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the small and the large intestine are part of the digestive system the small intestine longer than the large intestine and every
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5 + n = large intestine  :-):-):-)
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Question in the photo
melisa1 [442]

Answer:

14

Step-by-step explanation:

To make this not a function, the same input would need to go to two different outputs

so the input would need to be 5 ,17, or 14

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3 years ago
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In a cafeteria, there are the same number of 3-legged stools as there are of 4-legged chairs. If stools have 120 fewer legs than
olga_2 [115]
Imagine what x is quantity of stools. So
3x+120=4x
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3 years ago
A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees. Befor
Leya [2.2K]

Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 700, \sigma = 50.

The probability of a month having a PCE between $575 and $790 is the <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, hence:

X = 790:

Z = \frac{X - \mu}{\sigma}

Z = \frac{790 - 700}{50}

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

X = 575:

Z = \frac{X - \mu}{\sigma}

Z = \frac{575 - 700}{50}

Z = -2.5

Z = -2.5 has a p-value of 0.0062.

0.9641 - 0.0062 = 0.9579.

0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

3 0
1 year ago
Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.
DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548  approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: 26\times 26\times 26 \times 10 \times 10 \times10

26^3\times 10^3=17576000

The numbers of ways in which two letters followed by four digits can be chosen is: 26\times 26\times 10 \times 10 \times 10 \times10

26^2\times 10^4=6760000

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: 26^2=676

The probability that a randomly chosen plate contains the number 2222 is:

\frac{676}{24336000} \approx 0.000028

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

26\times 1\times10^3+1\times 26\times10^3

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

1\times 10^4

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

1\times 10^4+26\times 1\times10^3+1\times 26\times10^3

62000

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

\frac{62000}{24336000} \approx 0.002548

7 0
3 years ago
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