A recipe for individual chocolate hazelnut tarts calls for ½ cup of hazelnuts per tart. If 1 cup of hazelnuts weighs 4 ounces, w
1 answer:
Answer:
The amount is sufficient to make 75 tarts.
Step-by-step explanation:
We have been given that a recipe for individual chocolate hazelnut tarts calls for ½ cup of hazelnuts per tart and 1 cup of hazelnuts weighs 4 ounces.
The half cup of hazelnuts will weigh 2 ounces .
1 kg equals 35.274 ounces.
Since each tart needs ½ cup of hazelnuts and half cup of hazelnuts will weigh 2 ounces, so we will divide 176.37 ounces by 2 to find number of tarts.
Since we can make 88 tarts from 5 kg hazelnuts, therefore, the 5-kilogram bag of hazelnuts be sufficient to make 75 tarts.
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Answer:
This really isn't a question. I can help you out with writing one.
John randomly selected 45% of seventh grade students in his school and asked them what their favorite time to play sports was. Of the students surveyed, 51 prefer to play sports after school. Based on this data, what is the most reasonable prediction of seventh graders in his entire grade?
Step-by-step explanation:
To answer my question, you would set up a proportion.
<-- the answer to that question will will be 113.
Answer:
Since a/2⁽ⁿ ⁺ ¹⁾b < a/2ⁿb, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.
Step-by-step explanation:
Let a/b be the rational number in its simplest form. If we divide a/b by 2, we get another rational number a/2b. a/2b < a/b. If we divide a/2b we have a/2b ÷ 2 = a/4b = a/2²b. So, for a given rational number a/b divided by 2, n times, we have our new number c = a/2ⁿb where n ≥ 1
Since = a/(2^∞)b = a/b × 1/∞ = a/b × 0 = 0, the sequence converges.
Now for each successive division by 2, a/2⁽ⁿ ⁺ ¹⁾b < a/2ⁿb and
a/2⁽ⁿ ⁺ ¹⁾b/a/2ⁿb = 1/2, so the next number is always half the previous number.
So, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.