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Arlecino [84]
3 years ago
11

How can 2,100,000,000 be written in scientific notation?

Mathematics
1 answer:
IgorLugansk [536]3 years ago
6 0

Answer:

for this the correct answer is option 3

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What do you do to find m?
lesya692 [45]
Ok here is how to solve these I hope this helps
7       =      2-5m
+5m           +5m
7+5m =     2
-7               -7
5m      =    -5
/5              /5
m=-1


<span>-Hope this helps :)</span>
8 0
3 years ago
Rewrite the following in the form log(c).<br> log(8) – log(2)
jek_recluse [69]

Answer:

log(c). log(8) - log(2)

Step-by-step explanation:

8 0
3 years ago
Translate this verbal expression to an algebraic expression.
puteri [66]
C. 2(x+10) is the correct answer
3 0
3 years ago
Read 2 more answers
The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard
bulgar [2K]

We have been given that the lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard  deviation is 2.4 years. We are asked to find the probability of a lion living longer than 10.1 years using empirical rule.

First of all, we will find the z-score corresponding to sample score 10.1.

z=\frac{x-\mu}{\sigma}, where,

z = z-score,

x = Random sample score,

\mu = Mean

\sigma = Standard deviation.

z=\frac{10.1-12.5}{2.4}

z=\frac{-2.4}{2.4}

z=-1

Since z-score of 10.1 is -1. Now we need to find area under curve that is below one standard deviation from mean.

We know that approximately 68% of data points lie between one standard deviation from mean.

We also know that 50% of data points are above mean and 50% of data points are below mean.

To find the probability of a data point with z-score -1, we will subtract half of 68% from 50%.

\frac{68\%}{2}=34\%

50\%-34\%=16\%

Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.

3 0
3 years ago
Please please help me!!
wolverine [178]
What do u need help with??
3 0
3 years ago
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