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natali 33 [55]
3 years ago
7

What is the range for this set of data? 7, 15, 12A. 3B. 5C. 8D. 12​

Mathematics
2 answers:
UNO [17]3 years ago
7 0

Answer:

8 is your answer

Step-by-step explanation:

Hope this helps sorry if I'm wrong

umka2103 [35]3 years ago
4 0

Answer:

8

Step-by-step explanation:

I'm doing the quiz rn.

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In the table below, x represents miles traveled and y represents the cost to travel by train.
Anastasy [175]

Answer:

The second equation is correct.

y=2.25x + 4

Step-by-step explanation:

We can systematically eliminate the others.

For equation 3 and 4, 6.5*2 and 12*x is already bigger than 8 of the first set of numbers.

For equation 1, 2*x+8.5 is also bigger than 8 from the first set.

Plugging in the values to equation 2, we always get the correct cost.

5 0
3 years ago
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30 POINTS,NEED HELP ASAP !!!
zaharov [31]

Answer: OPTION A.

Step-by-step explanation:

You can observe that in the figure CDEF the vertices are:

C(-2,-1),\ D(-2,0),\ E(2,2)\ and\ F(2,1)

And in the figure C'D'E'F'  the vertices are:

C'(-8,-4),\ D'(-8,0),\ E'(8,8)\ and\ F'(8,4)

For this case, you can divide any coordinate of any vertex of the figure C'D'E'F' by any coordinate of any vertex of the figure CDEF:

For C'(-8,-4) and C(-2,-1):

\frac{-8}{-2}=4\\\\\frac{-4}{-1}=4

Let's choose another vertex. For E'(8,8) and E(2,2):

\frac{8}{2}=4\\\\\frac{8}{2}=4

You can observe that the coordinates of C' are obtained by multiplying each coordinate of C by 4 and the the coordinates of E' are also obtained by multiplying each coordinate of E by 4.

Therefore, the rule that yields the dilation of the figure CDEF centered at the origin is:

(x, y)→(4x, 4y)

7 0
3 years ago
Read 2 more answers
Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
3 years ago
Erica invests $10,000 at 5% interest compounded annually.
grigory [225]
A=P(1+i)^{n}
A=10000(1+15%)^{3}
=11576.25
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3 years ago
PLEASE HELP AND PLEASE DON'T GUESS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Sindrei [870]
The 8th term is 33 and I don't think any of the options are there
7 0
3 years ago
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