Step-by-step solution-:

The Given that:

**8.6~oz** of mixed nuts that contains **20%** peanuts:

Let be amount of peanuts be **a** :

By using the formula of **Percent Proportion**:

The percent proportion representing that **x** is **p%** of **y** is given by:

\frac{x}{y}=\frac{P}{100}

Substitute the **y=8.6 oz** and **p=20%** into the **Percent Proportion**:

\frac{a}{\textcolor{primary}{8.6}}=\frac{\textcolor{secondary}{20}}{100}

After Cross multiplication:

100a=8.6\times20

100a=172=>a=\frac{172}{100}=>a=1.72

Let **b** be the mass of mixed nuts that contains **65%** peanuts:

Similarly,

By using the formula of **Percent Proportion**:

\frac{x}{y}=\frac{P}{100}

Substitute the **y=b oz** and **p=65** into the **Percent Proportion**:

\frac{x}{\textcolor{primary}{b}}=\frac{\textcolor{secondary}{65}}{100}

After solving we get,

x=\frac{13}{20}b

The total amount of peanuts=**(1.72+(13/20)b) oz**:

The Total amount will be original** 8.6 oz** plus the added amount of peanuts

The total amount of nuts =**(8.6+b) oz**:

Substitute the **y=(8.6+b) oz, x=(1.72+\13/20b)oz** and **p=50** into the **Percent Proportion**:

\frac{\textcolor{tertiary}{(1.72+\frac{13}{20}b)}}{\textcolor{primary}{(8.6+b)}}=\frac{\textcolor{secondary}{50}}{100}

After solving we get **b**:

b=17.2

Therefore, The value of **b** is **17.2 oz**:

The Ry was added to **17.2 oz** of mixed nu that contain **65%** peanuts.

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