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Minchanka [31]
3 years ago
7

URGENTTT 3+3 x 7^2 - 9 x 10

Mathematics
1 answer:
charle [14.2K]3 years ago
8 0

Answer: 60

Step-by-step explanation:

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1. Determine the median of this data set.
Simora [160]
1) When finding a median, order your numerical list from smallest to larger, as so: 2,2,3,4,5,6,8,9.
2) Then start eliminating numbers from each side of the order; (The bolded numbers will be removed.)
    2,2,3,4,5,6,8,9
   
2,3,4,5,6,8,9
    2
,3,4,5,6,8
    3,4,5,6,8
    
3,4,5,6
    4,5,6
    
4,5
3) Since there is an even number of digits in your list, you will have to add the last two digits and then divide them by two. 4 and 5 are the last numbers standing, (4+5=9)(9/2=4.5) and therefore your answer, would be 4.5, or B.
8 0
3 years ago
Question 7: 10 pts
Natalija [7]

Answer:

3 divid by 10 is 3 so i guses you have a 3 percent chance at getteing a green marebel

Step-by-step explanation:

6 0
2 years ago
Expand and simplify this equation plz, 4(3x+5) - 5(8x+2)
omeli [17]

4(3x + 5) - 5(8x + 2) \\ 12x + 20 - 40x - 10 \\ 12 x - 40x + 20 - 10 \\  - 28x + 10

7 0
2 years ago
An SRS of 450 450 high school seniors gained an average of ¯ x = 20 x¯=20 points in their second attempt at the SAT Mathematics
maria [59]

Answer: (15.47, 24.53)

Step-by-step explanation:

We know that the confidence interval for population mean is given by :_

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

, where n= sample size.

\sigma = standard deviation.

\overline{x}= sample mean.

z*= Critical value.

Given : n= 450

\overline{x}=20

\sigma=49

Critical value for 95% confidence = z*=1.96     [From z-value table]

Then, the 95% confidence interval will be :-

20\pm (1.96)\dfrac{49}{\sqrt{450}}

\approx 20\pm (1.96)(2.31)

\approx 20\pm 4.53

=(20-4.53,\ 20+4.53)=(15.47,\ 24.53)

Hence, the 95% confidence interval for the mean change in score μ μ in the population of all high school seniors. : (15.47, 24.53)

3 0
3 years ago
Find the sum of the convergent series 7 0.7 0.007
mafiozo [28]
7+0.7+0.007+\cdots=7\left(1+\dfrac1{10}+\dfrac1{100}+\cdots\right)

Let S_n denote the nth partial sum of the series, i.e.

S_n=1+\dfrac1{10}+\dfrac1{100}+\cdots+\dfrac1{100^n}

Then

\dfrac1{10}S_n=\dfrac1{10}+\dfrac1{100}+\dfrac1{1000}+\dfrac1{10^{n+1}}

and subtracting from S_n we get

S_n-\dfrac1{10}S_n=\dfrac9{10}S_n=1-\dfrac1{10^{n+1}}
\implies S_n=\dfrac{10}9\left(1-\dfrac1{10^{n+1}}\right)

As n\to\infty, the exponential term vanishes, leaving us with

\displaystyle\lim_{n\to\infty}S_n=\dfrac{10}9

and so

7+0.7+0.007+\cdots=7.777\ldots=\dfrac{70}9
3 0
3 years ago
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