Find the -5/8 -- 3/10

1 answer:

6 0

I’m not for sure of the answer but i do know that the two subtractions in the middle would turn into a + and i think you add the -5/8 + 3/10

You might be interested in

How do I do number 7 please help me with

riadik2000 [5.3K]

There is a Khana Academy video that explains everything you have to do and it’s the same page that you’re on

3 0

2 years ago

Please help me just number 8

Rufina [12.5K]

Answer:

a. 3.9714285714 b.1.404040404

Step-by-step explanation:

13.9/3.5

13.9/9.9

I do not know if this is correct, but give it a try.

4 0

3 years ago

Read 2 more answers

The temperature is -3 F at 7:00 am During the next 4 hours the temperature increases 21 F. What is the temperature at 11:00am ?

telo118 [61]

Now what you want to do is 21 + (-3). Since this is the same as saying 21 - 3, the answer is 18. Therefore, the answer is 18°F.

4 0

3 years ago

There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit

Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a $\frac{3}{11}$ chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is $\frac{8}{11}$ , and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

$\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}$

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is $\frac{2}{11}$ , and the probability of picking the second red ball is $\frac{1}{10}$ and the probability of picking the green ball is $\frac{1}{9}$ . We want to multiply thm again, as per the multiplication rule like the last problem.