Is this the correct way to solve these math problems and are the answers correct.

Directions:Solve the following applications.
Science. Jose rode his trail bike for 10miles. Two-thirds of the distance was over a mountain trail. How long is tehe mountain trail?

2/3 (x)10= 6 and 2/3 or 6.666666666667

Statistics. IN a weekend baseball tournament, Joel ha 4hits in 13times at bat. THat is , he hit safely 4/13 of the time. Write the decimal equivalent for Joel's hitting , round to the three decimal places.(that number is Joels batting average.

4divided by 13 = .307 or 3.07

I'm not Bob Pursley but I can answer your question. 6&2/3 is correct. If I changed that to a decimal, I would round it off at 6.67 miles.

For the batting average, when I divided 4/13, I obtained 0.30769. The instructions are to round to three places; therefore, I would round to 0.308.

Thank you for your help

Can someone show me how to work this problem. It is stressing me out how you came up with the answer. The one about the mountain trail bike?

Sure! I'd be happy to explain how to solve the problem about the mountain trail bike.

The problem statement tells us that Jose rode his trail bike for 10 miles, and two-thirds of the distance was over a mountain trail. We need to find the length of the mountain trail.

To solve this, we can use the concept of proportions. We know that two-thirds of the total distance is the length of the mountain trail. Let's call the length of the mountain trail "x".

The proportion would look like this:

(2/3) / 10 = x / 1

To solve this proportion, we can cross-multiply and solve for x:

(2/3) * 1 = 10 * x

2/3 = 10x

Now we isolate x by dividing both sides of the equation by 10:

(2/3) / 10 = x

2/30 = x

Simplifying the fraction 2/30, we get:

1/15 = x

So, the length of the mountain trail is 1/15 of the total distance, which is approximately 0.067 miles.

Therefore, the answer is that the mountain trail is approximately 0.067 miles long.

I hope this explanation helps! Let me know if you have any other questions.