3(x-4)+2(×+1) I need help

pls solve quickly!: It took a boat 2 hours to reach town A going upstream. The way back was 1h20 min. What is the speed of the b

goldenfox [79]

Answer:

15 mph

Step-by-step explanation:

Given: Boat took 2 hours to reach Town A going upstream.

Speed of stream= 3 mph

Time taken to reach back home= 1 hours 20 minutes

Lets assume distance covered one side be "d" and speed of boat in still water be "s".

∴ Speed of boat in upstream= $(s-3) \ mph$

Speed of boat in downstream= $(s+3)\ mph$

Also converting into fraction of time taken to reach back home.

Remember; 1 hour= 60 minutes

∴ Time taken to reach back home= $60+20= 80\ minutes$

Converting time given into fraction= $\frac{80\ minutes}{60\ minutes} = \frac{4}{3} \ hours$

hence, Time taken to reach back home is $\frac{4}{3} \ hours$

Now forming equation of boat travelling upstream and downstream, considering distance remain constant.

We know, $Distance= speed \times time$

⇒ $(s-3)\times 2= (s+3)\times \frac{4}{3}$

Using distributive property of multiplication

⇒ $2s-6= \frac{4}{3}s +4$

subtracting both side by $\frac{4}{3} s$

⇒ $2s-\frac{4}{3} s-6= 4$

Adding both side by 6

⇒ $2s-\frac{4}{3} s= 10$

taking LCD as 3

⇒ $\frac{2}{3} s= 10$

Multiplying both side by $\frac{3}{2}$

⇒ $s= \frac{3}{2} \times 10$

∴s= 15 mph

Hence, 15 mph is the speed of the boat in still water.

8 0

3 years ago

It is in the attachment.

Setler79 [48]

Answer:

you would have to pay $43.20

Step-by-step explanation:

15+32.50=47.50

47.50x.08=3.80

47-3.80=43.20

3 0

3 years ago

Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5