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m_a_m_a [10]
3 years ago
6

PLZ PLZ PLZ HELP MEEEE

Mathematics
2 answers:
Citrus2011 [14]3 years ago
5 0
I Believe the answer is 16 and 4
quester [9]3 years ago
3 0

Answer:

open comment

please mark me brilliant

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Please solve this question faaaast​
Kazeer [188]

Answer:

300-400

Step-by-step explanation:

The first step is finding the total of the data we have. So, we take 5 + 10 + 15 + 20 + 25 + 15 + 10 which equals 80.

The median is the middle point of all the data. If it's an odd number, you can calculate the median with the equation (n+1) / 2, plugging in the total amount of data for n.

When it's an even number though, there is no one middle point since the data splits evenly in 2, so we have to use 2 equations: n/2 & (n/2) + 1. Then, we find the average of those two data points. (Although, you don't have to do that for this particular question).

When we plug 80 in for n in both of these equations, we get 40 and 41.

To where this is in the question, we have to count up from the bottom of the chart. 1-5 is below 100, 6-15 is 100-200, 16-30 is 200-300, and 31-50 is 300-400.

Since 40 and 41 are between 31 and 50, the answer is 300-400.

Hope this helps! :)

6 0
2 years ago
If you flip three fair coins what is probability that you'll get all three heads
Murljashka [212]
1/2 times 1/2 times 1/2 is 1/8
Probability of head times the probability of head times the probability of head


So 1/8

4 0
4 years ago
Read 2 more answers
If V = 1/3 BH, what is h expressed in terms of B and V?
Margaret [11]

V=(1/3)(B)(h)

multiply both sides by 3

  • 3V=Bh

divide both sides by B

  • 3V/B=h
8 0
3 years ago
In 2010, the Census Bureau estimated the proportion of all Americans who own their homes to be 0.669. An urban economist wants t
Alexxandr [17]

Answer:

i)

Sample size making use of the Census Bureau: 1,499 American adults.

Sample size without making use of the Census Bureau: 1,692 American adults

ii)

71

Step-by-step explanation:

i)

The sample size n in Simple Random Sampling is given by

\bf n=\frac{z^2p(1-p)}{e^2}

where  

<em>z = 1.645 is the critical value for a 90% confidence level </em><em>(*) </em>

<em>p= 0.669 is the population proportion given by the Census  </em>

<em>e = 0.02 is the margin of error </em>

so  

\bf n=\frac{(1.645)^2*0.669*0.331}{0.02^2}=1,498.05\approx 1,499

rounded up to the nearest integer.

(*)This is a point z such that the area under the Normal curve N(0,1) 1nside the interval [-z, z] equals 90% = 0.9

<em>It can be obtained with tables or in Excel or OpenOffice Calc with </em>

<em>NORMSINV(0.95) </em>

<em> </em>

If she ignores the Census estimate, the she has to take the largest sample possible that meets the requirements.

Let's show it is obtained when p = 0.5

As we said, the sample size n is

\bf n=\frac{z^2p(1-p)}{e^2}

where  

e = 0.02 is the error proportion  

z = 1.645

hence

\bf n=\frac{(1.645)^2p(1-p)}{(0.02)^2}=6765.0625p(1-p)=6765.0625p-6765.0625p^2

taking the <em>derivative</em> with respect to p, we get

n'(p)=6765.0625-2*6765.0625p

and  

n'(p) = 0 when p=0.5

By taking the second derivative we see n''(p)<0, so p=0.5 is a maximum of n

<em>This means that if we set p=0.5, we get the maximum sample size for the confidence level required for the proportion error 0.02 </em>

Replacing p with 0.5 in the formula for the sample size we get

\bf n=6765.0625*0.5-6765.0625(0.5)^2=1691.27\approx 1,692

rounded to the nearest integer.

ii)

When we do not have a proportion but a variable whose approximate standard deviation s is known, then the sample size n in Simple Random Sampling is given by

\bf n=\frac{z^2s^2}{e^2}

where  

<em>z = 2.241 is the critical value for a 95% confidence level </em><em>(*) </em>

<em>s = 7.5 is the estimated population standard deviation </em>

<em>e = 2 hours is the margin of error </em>

so  

\bf n=\frac{z^2s^2}{e^2}=\frac{(2.241)^2(7.5)^2}{(2)^2}=70.62\approx 71

(*)This is a point z such that the area under the Normal curve N(0,1) inside the interval [-z, z] equals 95% = 0.95

<em>It</em> <em>can be obtained in Excel or OpenOffice Calc with </em>

<em>NORMSINV(0.9875) </em>

5 0
4 years ago
What is the direct linear variation equation for the relationship?
defon
\bf \begin{array}{cccccclllll}&#10;\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\&#10;\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\&#10;y&=&{{ K}}&\cdot&x&#10;\\ \\&#10;&&  y={{ K }}x&#10;\end{array}&#10;\\ \quad \\&#10;&#10;\begin{cases}&#10;y=4\\&#10;x=12&#10;\end{cases}\implies y=Kx\implies 4=K\cdot 12

solve for "K", to see what "K" is, or the "constant of variation",
once you found K, plug it back in y=Kx, and that's the equation
3 0
3 years ago
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