There are \({6 \choose 3 }= 20\) ways to choose 3 boys for the team
There is \({8 \choose 4} = 70\) ways to choose 4 girls for the team
Thus, there are \(20 \times 70 = \color{brown}\boxed{1400}\) ways to choose 3 boys and 4 girls for the team.
This is evidently from a test. A test is made to test YOUR capabilities, not ours. You should not be asking these questions, and we should not be answering these questions.
Parity Rules:
pos x pos = pos
pos x neg = neg
neg x pos = neg
neg x neg = pos
Both \(1\over2\) and \(12\) are positive, so the product must be \(\color{brown}\boxed{\text {positive}}\)
We know that \(\angle BCE =30\). We know that \(BC\) and \(CE\) are the same length, so\(\triangle{CEB}\) is isosceles. This means that \(\angle CBE=\color{brown}\boxed{75^{\circ}}\)
Part A: \({1\over2}\times{12}=\color{brown}\boxed{6}\)
Part B: \(\color{brown}\boxed{\text{positive}}\)
763 thousanths is equal to 0.763, or \(\color{brown}\boxed{7.63 \times {10^{-1}}}\)
The smallest odd integer is \(301\), and the largest one is \(599\). There are \(150\)numbers in this series, so the sum is \(150 \times 450 = \color{brown}\boxed{67,500}\)
The max height occurs at \(-b/2a\). This means that the x-axis of hte peak is \({-60\over-32} = 1 {7\over8}\)
Subsitute \(1 {7\over8}\) for \(t\), and you find that the max height is \(\color{brown}\boxed{77.25}\)
The smallest number possible is 1, and the largest is 15.
Of these, all \(\color{brown}\boxed{15}\) numbers are possible.
Look, we answer questions on this forum. We don't look up answer keys for you.
