Look at photo for question. Do number 3 A-D?

Mathematics

1 answer:

Digiron [165]3 years ago

8 0

Answer:

A: 6, 12, 18, 24, 30

10, 20, 30, 40

B: 30

C: 5/6= 25/30 and 7/10= 21/30

D: 25/30>21/30 so 5/6>7/10

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3 letters without replacement 4 letters A B C D how many ways can this be done if the order of the choices matters

Veseljchak [2.6K]

Answer:

Since the order of choice matters, we will permute the values. a bit more explanation for this:

If the order of choice did NOT matter, ABC and BCA will be counted as one since order of choice does NOT matter

Since order of choice does matter, ABC , BCA and CAB are all different possibilities for the arrangement of the same 3 letters

Since we have 3 slots:

___ ___ ___

Now, for the first slot. You can out either one if the 4 alphabets in the first slot since no slot has been used as of now

So:

_4_ ___ ___

**Keep in mind that the 4 is the possible number of values this slot can have**

Now that one slot has been used, one of the 4 alphabets has been used and since we are not allowed to repeat the same alphabets, we are left with 3 more alphabets

we can put any one of the 3 alphabets in this second slot, Hence:

_4_ _3_ ___

Now that 2 of the 4 alphabets have been used, we are left with only 2 alphabets, so there are only 2 possible alphabets for slot 3

Therefore:

_4_ _3_ _2_

Now that we know the possible alphabets for all 3 slots, we will multiply them with each other to get the total possible number of 3 - alphabet words we can make with 4 alphabets

Total possible words = 4 * 3 * 2

Total possible words = 24

We could've used the formula for Permutation as well

8 0

3 years ago

By what number we multiply -7/16 so that the product may be 46/28

Aliun [14]

Answer:

The eq will be=-7/16 × x=46/28

x=46/28 ×-16/7

x=-192/49

3 0