A 4.0 cm tall object is placed 8.3 cm in front of a concave mirror with a focal length of 15.2 cm. Calculate the image distance and image height.

The given values are:
Object height, $h_o = 4.0 \text{ cm}$
Object distance, $u = 8.3 \text{ cm}$
Focal length of the concave mirror, $f = 15.2 \text{ cm}$

We need to find the image distance ($v$) and the image size ($h_i$).
The mirror formula is:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

The magnification formula is:

$$M = \frac{h_i}{h_o} = -\frac{v}{u}$$

For a concave mirror, the focal length $f$ is positive. The object distance $u$ is positive for a real object.

Rearrange the mirror formula to solve for $v$:

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$

Substitute the given values:

$$\frac{1}{v} = \frac{1}{15.2 \text{ cm}} - \frac{1}{8.3 \text{ cm}}$$ $$\frac{1}{v} = 0.065789 \text{ cm}^{-1} - 0.120482 \text{ cm}^{-1}$$ $$\frac{1}{v} = -0.054693 \text{ cm}^{-1}$$ $$v = \frac{1}{-0.054693 \text{ cm}^{-1}}$$ $$v \approx -18.28 \text{ cm}$$

The negative sign indicates that the image is virtual and formed behind the mirror.

Use the magnification formula:

$$\frac{h_i}{h_o} = -\frac{v}{u}$$

Rearrange to solve for $h_i$:

$$h_i = -h_o \left(\frac{v}{u}\right)$$

Substitute the known values:

$$h_i = -(4.0 \text{ cm}) \left(\frac{-18.28 \text{ cm}}{8.3 \text{ cm}}\right)$$ $$h_i = (4.0 \text{ cm}) \left(\frac{18.28}{8.3}\right)$$ $$h_i = (4.0 \text{ cm}) \times 2.2024$$ $$h_i \approx 8.81 \text{ cm}$$

The positive sign indicates that the image is upright.

The image distance is $\boxed{\text{-18.3 cm}}$ and the image size is $\boxed{\text{8.81 cm}}$.