All categories

3 years ago

Your cousin gave you 27 dollars with which to buy a present. this covered 9/10 of the cost. how much did the present cost

2 answers:

4 0

The other answer is working the wrong way.

27 dollars is ONLY 9/10 of the cost.

Which means we have to do the following:

27/9=3
3*10 = 30

The total cost of the present, was 30 dollars.

yaroslaw [1]3 years ago

3 0

27/10 is 2.7
2.7*9 gets you $24.3

:)

You might be interested in

The kitchen-tile installer has 20 green, 14 beige, and 16 white tiles in a box. What is the probability of picking a beige tile

Stels [109]

Answer: P = 0.28

Step-by-step explanation:

If we pick it at random, the probability of selecting each tile must be equal. this means that the probability of picking a tile of a giving color is equal to the number of the tiles of that color divided the total number of tiles.

We have 14 beige tiles and 20 + 14 + 16 = 50 tiles in total.

Then the probability of piking a beige tile is:

P = 14/50 = 0.28

8 0

2 years ago

Simplify 38 over 49 what’s the anwswer

Katyanochek1 [597]

Answer: 38 over 49 can be written as 38/49. When divided, you will get 0.77551020408. It takes lots of regrouping, but it ends eventually.

If you are reducing it as a fraction, it is not possible because there is no other number to reduce it, so it will be in it's simplest form already.

8 0

2 years ago

Read 2 more answers

A car manufacturer sent out survey cards to owners who had purchased new cars. The survey card only had boxes to check for repli

pantera1 [17]

Answer: I think the answer is B. I don't fully understand the question but it seems like that would be the answer. You might double check though.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Use mathematical induction to show that 4^n ≡ 3n+1 (mod 9) for all n equal to or greater than 0

cestrela7 [59]

When $n=0$ , you have

$4^0=1\equiv3(0)+1=1\mod9$

Now assume this is true for $n=k$ , i.e.

$4^k\equiv3k+1\mod9$

and under this hypothesis show that it's also true for $n=k+1$ . You have

$4^k\equiv3k+1\mod9$
$4\equiv4\mod9$
$\implies 4\times4^k\equiv4(3k+1)\mod9$
$\implies 4^{k+1}\equiv12k+4\mod9$

In other words, there exists $M$ such that

$4^{k+1}=9M+12k+4$

Rewriting, you have

$4^{k+1}=9M+9k+3k+4$
$4^{k+1}=9(M+k)+3k+3+1$
$4^{k+1}=9(M+k)+3(k+1)+1$

and this is equivalent to $3(k+1)+1$ modulo 9, as desired.

3 0

3 years ago

M plus the product of 6 and n

irina [24]

Answer:

m+6n

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers