All categories

3 years ago

In the data set {3, 7, 1, 4, 5, 7} what is the mean absolute deviation

1 answer:

6 0

The answer would be 11 because you add them all up the divide by how many numbers you have witch is 6 and you get 4.5. Then you subtract 4.5 from all the numbers of the set and take the absolute value then add them all up again and that how you would get 11.hope I wasn’t to late!

You might be interested in

What is the value of 4^-3 in fractions

natima [27]

It will be 1/4^3 then 1/64

8 0

3 years ago

HELP ILL MARK BRAINLEST

ivolga24 [154]

Answer:

Step-by-step explanation:

The answer is C, as as any value of m will make the equation true.

Such as if we used 2 as a substite value for m.

$-6(2)-6+8(2)=-5+2(2)-1$

Lets simplify both equations:

$-6(2)-6+8(2)=-2$

$-5+2(2)-1=-2$

Now, if you want, you can also use any other value for m, but that is basically how I got C. If they equations didn't have the same answer, then it would be B. If it only worked for 2 numbers, then it would be A. Now, I tried this equation again with -3=m, but I still ended up with the same answers. So it wouldn't have been D.

8 0

4 years ago

Read 2 more answers

3 more than the product of 2 and a number x

Morgarella [4.7K]

2x+3. that is the answer because 2 and a number is 2x and 3 more than that would be adding 3

5 0

3 years ago

Tim read 70% of his book containing 230 pages how many pages does he have left

MakcuM [25]

Hello!

Considering the whole book as 100%, there’s 30% left.

Then, 30% of 230 = 0,3*230 = 69 pages.

Hope you’re all right! Stay safe!

3 0

4 years ago

WILL GIVE BRAINLIEST!!!

Mice21 [21]

Answer:

6 total

3 positive 2 digit numbers: 17, 52, 87

3 negative 2 digit numbers: -18, -53,-88

Step-by-step explanation:

We are given

x ≡ 3(mod7), x ≡ 2(mod5) and we want x to be two digits long.

x ≡ 3(mod7) means x-3=7k and

x ≡ 2(mod5) means x-2=5m where k and m are integers.

We need to find an x that satisfies both.

So some multiples of 7 are 7,14,21,28,35,42,49,56,63,70,77,84,91,98

Some multiples or 5 are

5,10 15,20,25,30 35,40,45,45,50,55,60,65,70

Now add 3 to first list for x=7k+3

10,17,24,31,38,45,52,59,66,73,80,87,94,101

Now add 2 to second list for x=5m+2

7,12 17,22,27,32 37,42,47,52,57,62,67,72

We only want to look at 2 digit numbers... just need to expand second list more:

Now add 2 to second list for x=5m+2

7,12,17,22,27,32,37,42,47,52,57,62,67,72,77,82,87,92,97,

That's all we need.

So let's write our lists and see the common numbers in them:

List 1: 10,17,24,31,38,45,52,59,66,73,80,87,94,101

List 2:

7,12,17,22,27,32,37,42,47,52,57,62,67,72,77,82,87,92,97

Common numbers: 17, 52, 87

Considering negative numbers as well:

x=7k+3:

3, -4, -11, -18, -25, -32,-39, -46, -53, -60, -67, -74, -81, -88, -95

x=5m+2:

2,-3,-8,-13,-18,-23,-28,-33,-38,-43,-48,-53,-58,-63,-68,-73,-78,-83,-88,-93,-98

Common 2digit numbers:

-18, -53,-88

7 0

3 years ago