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AleksandrR [38]
3 years ago
11

73,498 to the nearest thousand

Mathematics
2 answers:
KIM [24]3 years ago
6 0

Answer:

73,000

Step-by-step explanation:

brainliest?

(::)

Anestetic [448]3 years ago
6 0

Answer: 73000

Hope this helps, have a great day/night!

~IFoundJiminsLostJams~

Step-by-step explanation:

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Ineed help help me please
melamori03 [73]
The symbol ₁₂P₉ represents the permutations of 9  quantities out of 12.
By definition,
_{12}P_{9} =  \frac{12!}{(12-9)!} = \frac{12!}{3!}

From the calculator,
12! = 479,001,600
3! = 6

Therefore
₁₂P₉ = 479001600/6 = 79,833,600

Answer: 79,833,600
8 0
4 years ago
4<br> How many 3-cup servings are in 4 cups?<br> A. 1/2<br> B. 2/<br> C. 4<br> D. 12
anygoal [31]
This questions stated differently so I’m assuming it’s 4?
7 0
1 year ago
Please help
katrin2010 [14]
\dfrac{n^2+3n+2}{n^2+6n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n^2+2n+n+2}{n^2+4n+2n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n(n+2)+1(n+2)}{n(n+4)+2(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{(n+2)(n+1)}{(n+2)(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{n+1}{n+4}-\dfrac{2n}{n+4}=\dfrac{n+1-2n}{n+4}=\dfrac{1-n}{n+4}

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7 0
3 years ago
Out of 100 people sampled, 42 had kids. Based on this, construct a 99% confidence interval for the true population proportion of
maksim [4K]

Answer:

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

Out of 100 people sampled, 42 had kids.

This means that n = 100, \pi = \frac{42}{100} = 0.42

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.293

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 + 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.547

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

7 0
3 years ago
Please answer this correctly
Andrej [43]
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4 0
3 years ago
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