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Eddi Din [679]
3 years ago
13

Eli Whitney invented the the cotten gin

Mathematics
2 answers:
gtnhenbr [62]3 years ago
6 0

He did. Are you asking to make sure?

alekssr [168]3 years ago
4 0
The answer is True.
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Choose another name for plane s look at the picture for the full answer thank you for helping!
IgorC [24]

Answer:

c: plane ptm

Step-by-step explanation:

4 0
3 years ago
The linear equation x=5 would be a graph of a..
Lelechka [254]

Thw answer to your Problem would be A

7 0
3 years ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
What is the answer to this problem? 0.35 + 0.27 = ?
gavmur [86]
Have you never added 2-digit numbers?
35 +27 = (30 +20) +(5 +7) = 50 +12 = 62

35/100 +27/100 = (35 +27)/100 = 62/100
0.35 +0.27 = 0.62

The basic idea, taught in 3rd grade, is to line up the decimal points of the numbers, add enough zeros on the right to make the numbers all have the same number of digits to the right of the decimal point, then add in the usual way. The sum has the decimal point aligned with the rest of the numbers.

3 0
3 years ago
Read 2 more answers
Lisa tapes three 7inch by 9inch pieces of construction paper together to make a happy birthday sign for her mom. She uses a piec
USPshnik [31]
Depends on how the papers are connected. if they are connected with the 7cm side joining each other, use solution one. if they are connected with the 9cm side joining each other, use solution two.


perimeter of edges of sign= 2 [(9x3)+7]
= 68 inches

left over= 144-68
= 76 inches



OR


perimeter of edges of sign= 2[(7x3)+9]
= 60 inches

left over= 144-60
= 84 inches
7 0
3 years ago
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