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Mashutka [201]
3 years ago
15

Helena needs 3.5 cups of flour per loaf of bread and 2.5 cups of flour per batch of muffins. She also needs 0.75 cup of sugar pe

r loaf of bread and 0.75 cup of sugar per batch of muffins. Helena has 17 cups of flour and 4.5 cups of sugar available for baking.
Mathematics
2 answers:
STatiana [176]3 years ago
8 0

The <em><u>correct</u></em> answer is:

She can make 2 loaves of bread and 4 batches of muffins


Arlecino [84]3 years ago
4 0
From the given data, we can generate two equations with two unknowns. 

We let x = number of loaves of bread
            y = number of batches of muffins

For the equation of the flour requirement:
17 = 3.5x + 2.5y

<span>For the equation of the sugar requirement:
</span>4.5 = 0.75x + 0.75y

We evaluate the solutions by manipulating one of the equations into terms of the other. We use the first equation.We write x in terms of y.

x = (4.5/0.75) - y

Substitute the third equation to the second equation.

17 = (3.5((4.5/0.75)-y)) + 2.5y

Evaluating y and x, we have,

y = 4 and x = 2

Thus, from the amounts she has in hand, she can make 4 loaves of bread and 2 batches of muffins.
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madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

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