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solmaris [256]
3 years ago
13

How does a digit in the ten thousand place compere to a digit in the thousand place

Mathematics
1 answer:
Mademuasel [1]3 years ago
7 0
Ten thousand - 10,000

thousand literally just haves 1,000


if the twenty is before the three zeros, it would be 20,000 which is twenty thousand. its the same with other numbers. its super simple!
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Easy question!<br><br> what is -2/2 ?
Zielflug [23.3K]

Answer:

-1

Step-by-step explanation:

-2 divided by 2 = -1.

A negative divided by a positive equals a lesser negative.

4 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!!!!!!!!!!
MaRussiya [10]
The answer is the top answer

6 0
4 years ago
Estimate the difference 8 7/12 - 4 7/8 - 4/10
Gnesinka [82]
When you are told to estimate, you should be able to do the problem in your head.
8 7/12 is close to 8 1/2
4 7/8 is very nearly 5
4/10 is very nearly 1/2

8 1/2 - 5 - 1/2 First of all you can take the 1/2 away from 8 1/2
8 1/2 - 1/2 = 8
Now take away the 5


8 - 5 = 3

So the answer is very nearly
3 <<<<< answer.

Just for fun, I'll get the exact answer. Don't hand this one  in. It is 3 37/120
You learn to estimate because you are trying to guess whether or not you've entered everything correctly into your calculator. When writing a test, this is a very handy skill to have developed.

4 0
3 years ago
If Leah babysits for 7 hours this month, what the minimum number of hours she would have to
Nataliya [291]
1000$ hope this helps
6 0
3 years ago
A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz
marin [14]

Answer:

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean,\mu_xis calculated as:

\mu_x=\mu=204.9, \mu_x=sample \ mean

-The standard deviation,\sigma_x is calculated as:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968

8. For a random variable X.

-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181

-#The sample mean,\mu_x is calculated as:

\mu_x=\mu\\\\=75.9

#The sample standard deviation is calculated as follows:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

\mu_x=\mu=135.7

#Sample standard deviation:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452

7 0
3 years ago
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