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son4ous [18]
3 years ago
10

What is the sum of the first 28 positive odd numbers

Mathematics
1 answer:
lubasha [3.4K]3 years ago
5 0
The sum of the first 28 positive odd numbers is 724. 
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Find the 61st term of the following arithmetic sentence
Leno4ka [110]

Answer:

The 61st term would be 495

Step-by-step explanation:

In order to find the 61st term, you need to find the rule for the sequence. You'll notice that the numbers are going up by 8. Since there is no 0th term, the first term would be 8 more than the base.

15 - 8 = 7

Therefore the base is 7. As a result, you can model the sequence as:

An = 8n + 7

In which n is the number in the sequence. Now knowing this, we can plug in 61 and get the answer.

An = 8n + 7

An = 8(61) + 7

An = 488 + 7

An = 495

4 0
3 years ago
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How to set up 701 km -523 km 445 m
Citrus2011 [14]
Convert -  445m into 0.445km
<span>So it's 701-523.445 = 177.555km

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4 0
3 years ago
I still dont understand slope can someone help?
katen-ka-za [31]

m = slope\\(x1, y1) = coordinates of first point in the line\\(x2, y2) = coordinates of second point in the line

4 0
2 years ago
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Diana works in a building that is 130 feet tall. She is outside, looking up at the building at an angle of 37° from her feet to
AveGali [126]
Let x be her initial distance from the building, then tan 37 = 130/x
x = 130/tan 37 = 130/0.7536 = 172.5 feet

Let y be her distance from the building after moving forward, then
tan 40 = 130/y
y = 130/tan 40 = 130/0.8391 = 154.9

After moving forward, she is 172.5 - 154.9 = 17.6 ft closer.
4 0
3 years ago
A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poisson distribution to find the p
andreyandreev [35.5K]

Answer:

B) 0.894

Step-by-step explanation:

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 local animal rescue organization receives an average of 0.55 rescue calls per hour.

This means that \mu = 0.55

Probability that during a randomly selected hour, the organization will receive fewer than two calls.

P(X < 2) = P(X = 0) + P(X = 1)

In which

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

P(X = 0) = \frac{e^{-0.55}*(0.55)^{0}}{(0)!} = 0.577

P(X = 1) = \frac{e^{-0.55}*(0.55)^{1}}{(1)!} = 0.317

P(X < 2) = P(X = 0) + P(X = 1) = 0.577 + 0.317 = 0.894

5 0
3 years ago
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