Groq Compound Mini

The SI unit of force is the **newton (N)**.

Josh’s total out‑of‑pocket cost was - Purchase price: \$80,000 - Repairs: \$50,000 **Total cost** = \$80,000 + \$50,000 = **\$130,000**. The repairs increased the house’s value by 150 %, meaning the new value is \[ \text{New value} = \$80,000 \times (1 + 1.5) = \$80,000 \times 2.5 = \$200,000. \] Assuming he sells the house at this new value, his **profit** is \[ \text{Profit} = \text{Sale price} - \text{Total cost} = \$200,000 - \$130,000 = \boxed{\$70,000}. \]

Janet has - 16 eggs total - 3 eaten for breakfast - 4 used for muffins Remaining eggs = 16 − 3 − 4 = **9 eggs**. At $2 per egg, she earns 9 × $2 = **$18** per day at the farmers’ market.

The binary‑search algorithm runs in **Θ(log n)** time (base‑2 logarithm). - **Best case:** O(1) – the target is found at the first middle element examined. - **Average / worst case:** O(log n) – each step halves the remaining search interval. If implemented iteratively, it uses **O(1)** extra space; a recursive version uses **O(log n)** stack space.

Josh bought the house for \$80,000 and spent another \$50,000 on repairs, for a total investment of \[ 80{,}000 + 50{,}000 = \$130{,}000. \] The repairs “increased the value of the house by 150 %.” An increase of 150 % means the new value is the original value plus 150 % of that value: \[ \text{New value} = 80{,}000 \times (1 + 1.5) = 80{,}000 \times 2.5 = \$200{,}000. \] Profit is the final sale price minus the total amount he spent: \[ \text{Profit} = 200{,}000 - 130{,}000 = \boxed{\$70{,}000}. \]

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Binary search runs in **O(log n)** time (base 2). - **Worst‑case**: O(log n) – each step halves the remaining search interval until the target is found or the interval is empty. - **Average‑case**: Also O(log n) for a uniformly random target in a sorted array. - **Best‑case**: O(1) – if the target happens to be the middle element on the first probe. The space complexity is O(1) for the iterative version (or O(log n) for the recursive version due to the call stack).

Binary search runs in **O(log n)** time (base 2). - **Worst‑case**: O(log n) – each step halves the remaining search interval until the target is found or the interval is empty. - **Average‑case**: Also O(log n) for a uniformly random target in a sorted array. - **Best‑case**: O(1) – if the target happens to be the middle element on the first probe. The space complexity is O(1) for the iterative version (or O(log n) for the recursive version due to the call stack).

Josh’s total outlay was - Purchase price: $80,000 - Repairs: $50,000 So he spent **$130,000** in total. A 150 % increase means the house’s value became 250 % of the original price: \[ \text{New value} = 80{,}000 \times (1 + 1.5) = 80{,}000 \times 2.5 = \$200{,}000. \] Profit = Sale price – Total costs \[ \text{Profit} = 200{,}000 - 130{,}000 = \boxed{\$70{,}000}. \]

9.8 is the larger number.