\( 100=6 \int_{0.19}^{\text {xout }} d x \cdot 59.88 \) What is the value of \( x_{\text {out }} \) ?

To solve for \( x_{\text{out}} \), first, we need to simplify the integral equation. The integral of a constant \( 59.88 \) with respect to \( dx \) over the limits \( 0.19 \) to \( x_{\text{out}} \) is: \[ \int_{0.19}^{x_{\text{out}}} 59.88 \, dx = 59.88 \cdot (x_{\text{out}} - 0.19) \] Now, plugging this into the equation, we have: \[ 100 = 6 \cdot 59.88 \cdot (x_{\text{out}} - 0.19) \] Calculating \( 6 \cdot 59.88 \): \[ 6 \cdot 59.88 = 359.28 \] Now we substitute it back into the equation: \[ 100 = 359.28 \cdot (x_{\text{out}} - 0.19) \] Dividing both sides by \( 359.28 \): \[ \frac{100}{359.28} = x_{\text{out}} - 0.19 \] Calculating the left side gives: \[ x_{\text{out}} - 0.19 = 0.2785 \quad (\text{approximately}) \] Now we solve for \( x_{\text{out}} \): \[ x_{\text{out}} = 0.2785 + 0.19 \approx 0.4685 \] Thus, the value of \( x_{\text{out}} \) is approximately \( 0.4685 \).