Javier drove 45 miles, this represents 60% of his entire trip. What is the total number of miles in his trip

2 answers:

7 0

60% is the same thing as 6/10, or 3/5. So 45 miles is 3/5 of his trip. 45=3/5x, where x is the number of miles in his trip. Divide both sides by 3/5 and you should get 75 miles.

faust18 [17]3 years ago

5 0

Ratio and proportion

45mi/60%=x miles/100%
cross multiply
60x=4500
x=75 miles

hope this helps

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PLEASE HELP ASAP!!

Arada [10]

This is like the equivalent to a jar with 4 green balls and 6 white balls, where you are picking 3. (The 4 green balls signify the friends from kindergarten.)

You want to solve the probability that the first two balls are green and the third is white.
First draw --> 4 green out of 10 balls --> 4/10 = 2/5

Second draw --> 3 green out of 9 balls --> 3/9 = 1/3

Third draw --> 6 white out of 8 balls --> 6/8 = 3/4

2/5 x 1/3 x 3/4

= 6/60

= 1/10

so the answer is 1/10 (or 10%)

6 0

3 years ago

Read 2 more answers

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .