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aleksandr82 [10.1K]
3 years ago
15

What is the value of the number 8 in 4,382,087

Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
8 0
The value of the number is 80,000
Oxana [17]3 years ago
3 0
The value of 8 in 4,382,087 is ten thousands

4: millions
3: hundred thousands
8: ten thousands
2: thousands 
0: hundreds
8: tens
7: ones


hope this helps
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What is the equation of the line that passes<br> through the point (3, 4) and has a slope of -2/3
Rina8888 [55]

Answer:

y=-2/3x+6

Step-by-step explanation:

y-y1=m(x-x1)

y-4=-2/3(x-3)

y=-2/3x+6/3+4

y=-2/3x+2+4

y=-2/3x+6

6 0
2 years ago
A coin is tossed 200 times, if the probability of getting a head is 5/8, how many times tail is obtained in all in tossing the c
koban [17]

Answer:

You would get tails 75 times, or 75:200

Step-by-step explanation:

5:8 = x:200

multiply 8 by 25 to get 200

multiply 5 by 25 to get x (125)

Probability to get heads is 125 so subtract 200 by 125

200 - 125 = 75

6 0
3 years ago
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KatRina [158]

Answer:

B) 120

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
Let set C = {1, 2, 3, 4, 5, 6, 7, 8} and set D = {2, 4, 6, 8}.
g100num [7]
<span>1. If it is intersection then it SHOULD be included in both the sets right? Now we know that odd numbers from 1-100 but the second set are multiples of 5 from 50-150! So we mainly need to look for common numbers which are ODD and are a MULTIPLE OF 5 BETWEEN 50 - 100!! So A={51,53,57,59,61......99} B={55,60,65,70.......95} [We stop till 100 because set A has no such element] So what is A ∩ B here? A ∩ B = {All odd numbers and multiples of 5 between 50 - 100}

</span>
6 0
3 years ago
Situation D: Suppose that, in a one-minute period during an electrical storm, the number of lightning strikes on a radar antenna
Crank

Answer:

21.77% probability that the antenna will be struck exactly once during this time period.

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.

In this question:

\mu = 2.40

Find the probability that the antenna will be struck exactly once during this time period.

This is P(X = 1).

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

P(X = 1) = \frac{e^{-2.40}*2.40^{1}}{(1)!} = 0.2177

21.77% probability that the antenna will be struck exactly once during this time period.

3 0
3 years ago
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