You mix 2 and one half

Given that Of 450 college students, 110 are enrolled in math, 205 are enrolled in English, and 50 are enrolled in both. If a student is selected at random

the probability

(a) The student is enrolled in mathematics= $\frac{110}{450} =0.244\\$

(b) The student is enrolled in English.= $\frac{205}{450} =0.456\\$

(c) The student is enrolled in both.= $\frac{50}{450} =0.111\\$

(d) The student is enrolled in mathematics or English.= $\frac{110+205-50}{450} =0.589\\$

(e) The student is enrolled in English but not in mathematics.

= $\frac{205-50}{450} =0.344\\$

(f) The student is not enrolled in English or is enrolled in mathematics.

= $\frac{450-205+110}{450} =0.7889\\$

8 0

3 years ago

Compute a · (b ✕ c), where a = i − 3j + k, b = 4i + j + k and c = 9i − j + 3k.

grandymaker [24]

$\mathbf a=\mathbf i-3\,\mathbf j+\mathbf k$
$\mathbf b=4\,\mathbf i+\mathbf j+\mathbf k$
$\mathbf c=9\,\mathbf i-\mathbf j+3\,\mathbf k$

$\mathbf b\times\mathbf c=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\4&1&1\\9&-1&3\end{vmatrix}=4\,\mathbf i-3\,\mathbf j-13\,\mathbf k$

$\mathbf a\cdot(\mathbf b\times\mathbf c)=(\mathbf i-3\,\mathbf j+\mathbf k)\cdot(2\,\mathbf i-3\,\mathbf j-13\,\mathbf k)=4+(-3)^2-13=0$

6 0

3 years ago