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irga5000 [103]
2 years ago
15

Someone help me please my friends don’t even know this question thank you

Mathematics
1 answer:
8090 [49]2 years ago
7 0

Step-by-step explanation:

from the question:R=x^2/y

where x=3.8*10^5

y=5.9*10^4

it then mean

R=(3.8*10^5)^2/5.9*10^4

=144400000000/59000

=2447457.63

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iren2701 [21]
Yes you would be correct
4 0
3 years ago
Harmon’s Jewelry Store receives a shipment of 500 jewelry pieces, but 15 pieces are damaged. At this rate, if Mrs. Harmon places
serious [3.7K]

Answer:

She should expect 165 jewelry pieces to be damaged.

Step-by-step explanation:

5,500/500 = 11

11 x 15 = 165

5 0
3 years ago
If two angles of a triangle have measures of 42° and 20°, what is the measure of the third? A. 118° B. 98° C. 108° D. 128°
dimaraw [331]

Answer: The text also states, “And her friends told her afterwards that she had spoken in a very loud voice, not shouting, just a very loud voice” (p.2). This proves that Celeste took a risk standing up to the teacher in a firm voice; her actions show that she believed the risk of getting in trouble was worth it to stop the Teacher’s unfair punishment.

Step-by-step explanation:

5 0
3 years ago
Helppp mee nowww I will give ya points and brianly est
asambeis [7]
X would be -5

Hope this helps :)
8 0
3 years ago
Estion
dalvyx [7]

Using the Poisson distribution, it is found that there is a 0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.

<h3>What is the Poisson distribution?</h3>

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

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

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • \mu is the mean in the given interval.

Considering the average of 15 birds every 2 hours during daylight hours, the mean for a 45-minute period is given by:

\mu = 15 \times \frac{45}{120} = 5.625

The probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours is given by:

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

In which:

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

P(X = 0) = \frac{e^{-5.625}(5.625)^{0}}{(0)!} = 0.004

P(X = 1) = \frac{e^{-5.625}(5.625)^{1}}{(1)!} = 0.02

P(X = 2) = \frac{e^{-5.625}(5.625)^{2}}{(2)!} = 0.057

P(X = 3) = \frac{e^{-5.625}(5.625)^{3}}{(3)!} = 0.107

P(X = 4) = \frac{e^{-5.625}(5.625)^{4}}{(4)!} = 0.150

P(X = 5) = \frac{e^{-5.625}(5.625)^{5}}{(5)!} = 0.169

Then:

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.004 + 0.02 + 0.057 + 0.107 + 0.15 + 0.169 = 0.507

0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

3 0
1 year ago
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