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ipn [44]
3 years ago
5

22.5 12.5 32 20 in least to greatest

Mathematics
2 answers:
pochemuha3 years ago
7 0
It would be 12.5, 20, 22.5, 32
Mrrafil [7]3 years ago
3 0
12.5, 20, 22.5, 32 is the way from least to greatest. Hope this helps.
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how many 4 digit numbers n have the property that the 3 digit number obtained by removing the leftmost digit is one ninth of n
Tom [10]

Answer:

  7

Step-by-step explanation:

We want to find the number 4-digit of positive integers n such that removing the thousands digit divides the number by 9.

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Let the thousands digit be 'd'. Then we want to find the integer solutions to ...

  n -1000d = n/9

  n -n/9 = 1000d . . . . . . add 1000d -n/9

  8n = 9000d . . . . . . . . multiply by 9

  n = 1125d . . . . . . . . . divide by 8

The values of d that will give a suitable 4-digit value of n are 1 through 7.

When d=8, n is 9000. Removing the 9 gives 0, not 1000.

When d=9, n is 10125, not a 4-digit number.

There are 7 4-digit numbers such that removing the thousands digit gives 1/9 of the number.

8 0
2 years ago
Is 1,2,3,4 a geometric sequence ?
Bad White [126]
No. It would be a arithmetic sequence
4 0
3 years ago
There are 40 batteries in 10 packs.
DaniilM [7]

:-):-):-):-):-)

Step-by-step explanation:

1) 40÷10

2) 40÷2

7 0
2 years ago
Melissa's flower shop got a shipment of 158 roses. She wants to make bouquets of 6 roses each. How many bouquets can Melissa mak
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158 and 6 she’ll havs 56 bouquets left
3 0
3 years ago
American kestrels live in More Mesa near your professor’s house. Suppose that there are 100 birds, of which a random sample of s
Ulleksa [173]

Answer:

P(X=3) = 0.2013

Step-by-step explanation:

After taking and tagging the first 20 birds, then we have a proportion of 20% of tagged birds within the population, that is

p=\frac{n}{N}

p=\frac{20}{100}=0.20

Supposing that the probability (0.20) of capturing a bird <em>remains constant</em>, we may resolve it as it follows:

Capturing 10 birds can be considered as a binomial experiment with a success probability p = 0.2

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p=0.2

x = 3

n=10

Then,

P(X=3)=10C3*0.2^{3}*0.8^{10-3}

P(X=x)=0.2013

Therefore, the probability that exactly 3 of the 10 birds in the new sample were previously tagged is 20.13%

5 0
3 years ago
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