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d1i1m1o1n [39]
3 years ago
9

ᶦᵐ ʲᵘˢᵗ ᵖᵘᵗᵗᶦⁿᵍ ᵃⁿ ᵃᶜᵃᵈᵉᵐᶦᶜ ᵠᵘᵉˢᵗᶦᵒⁿ ᶠᶦʳˢᵗ ˢᵒ ᶦᵗ ᵈᵒᵉˢⁿᵗ ᵍᵉᵗ ᵗᵃᵏᵉⁿ ᵈᵒʷⁿ

Mathematics
2 answers:
DerKrebs [107]3 years ago
5 0

Answer:

what's your name on Discord????

Step-by-step explanation:

have a good day

swat323 years ago
3 0

Answer:

sureee :)

Step-by-step explanation:

have a good day !!

~evita

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PLEASE HURRY!!!!!!!
nika2105 [10]

Answer:

The mid-point of the given line segment is:(5.5,-6.5)

Step-by-step explanation:

let us consider two points A and B with coordinates (a,b) and (c,d) respectively.

let C be the mid point of A and B then then the coordinates of C is given by (\frac{a+c}{2},\frac{b+d}{2}).

in the given question (a,b)=(9,-8) and (c,d)=(2,-5) so the mid point of these two points are (\frac{9+2}{2},\frac{-8-5}{2})=(5.5,-6.5)

Hence, the coordinates of the midpoint of the line segment with endpoints B(9,-8) and C(2,-5) is (5.5,-6.5).

8 0
4 years ago
Read 2 more answers
David earns $8 per hour. He works 40 hours each week. How much does he earn after 6 weeks
Vikki [24]
David will earn $1,920 in six weeks
8 x 40 = 320 a week
320 x 6 = 1,920 in six weeks
5 0
3 years ago
During the period of time that a local university takes phone-in registrations, calls come in
Zinaida [17]

Answer:

a) The expected number of calls in one hour is 30.

b) There is a 21.38% probability of three calls in five minutes.

c) There is an 8.2% probability of no calls in a five minute period.

Step-by-step explanation:

In problems that we only have the mean during a time period can be solved by the Poisson probability distribution.

Poisson probability distribution

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

a. What is the expected number of calls in one hour?

Calls come in at the rate of one each two minutes. There are 60 minutes in one hour. This means that the expected number of calls in one hour is 30.

b. What is the probability of three calls in five minutes?

Calls come in at the rate of one each two minutes. So in five minutes, 2.5 calls are expected, which means that \mu = 2.5. We want to find P(X = 3).

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 3) = \frac{e^{-2.5}*(2.5)^{3}}{(3)!} = 0.2138

There is a 21.38% probability of three calls in five minutes.

c. What is the probability of no calls in a five-minute period?

This is P(X = 0) with \mu = 2.5.

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-2.5}*(2.5)^{0}}{(0)!} = 0.0820

There is an 8.2% probability of no calls in a five minute period.

6 0
3 years ago
A study in which conditions are under the direct control of the investigator.
Andre45 [30]

Answer:

Experimental study

Step-by-step explanation:

3 0
3 years ago
Can someone please help me with this ? Dew in 5 mins .. I clicked the first answer on accident btw ..
Alexxx [7]

Answer:

B The equation is 4x = -48. Ben's point total is -12.

Step-by-step explanation:

Ben's total + Olive's total = -48

x + 3x = -48

4x = -48

x = -48 ÷ 4

Ben's total (x) = -12

3 0
3 years ago
Read 2 more answers
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