Tom knows that in his school 10 out of every 85 students are left-handed. There are 391 students in Toms school. How many are le
1 answer:
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Answer: 2/3ft or C
Step-by-step explanation: Solve for b in the area of a triangle equation, A=1/2bh. Multiply 1/2 by 2 on both sides to cancel it: A(2)=bh. Divide both sides by h: =b
A = 2/5
h = 6/5
b = ?
Next, we can plug the numbers into this equation: =b
Multiply 2 and 2/5: 4/5 / 6/5
Take the reciprocal of 6/5: 4/5 x 5/6 =
÷
=
2 ways: make a little divide tree based on how many times you can divide it. Like you have sqrt88 you know it can divide by 2. So its 2 and 44, and 44 can be broken to 2 and 22, then 2 and 11. 11 can’t be divided anymore.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.
Answer:
"2.82" seems to be the correct solution.
Step-by-step explanation:
As we know,
A bag (Bag A) contains,
Red balls = 3
Blue ball = 1
A second bag (Bag B) contains,
Red balls = 1
Blue ball = 1
Now,
⇒ P(Bag A, red ball) =
On substituting the values, we get
⇒ =
⇒ =
Similarly,
Probability of second bag will be:
⇒ P(Bag B) =
hence,
In the first bag (Bag A), the expected number of red balls will be:
⇒
On substituting the values, we get
⇒
⇒
⇒