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3 years ago

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Nell's mom makes chocolate milk with 30\text{ mL}30 mL30, space, m, L of chocolate syrup for every 222 ounces of milk. Nell's da

d adds 65\text{ mL}65 mL65, space, m, L of chocolate syrup for every 555 ounces of milk. Whose chocolate milk is more chocolatey? Choose 1 answer: Choose 1 answer: (Choice A) A Nell's mom's (Choice B) B Nell's dad's (Choice C) C The two chocolate milks are equally chocolatey. Stuck?Watch a video or use a hint.

2 answers:

klio [65]3 years ago

7 0

Answer:

The correct option is Option A) Nell's mom's

Step-by-step explanation:

Consider the provided information.

It is given that Nell's mom makes chocolate milk with 30 mL of chocolate syrup for every 2 ounces of milk.

This can be written as:

30 ml = 2 ounce of milk

Divide both the sides by 2.

15 ml = 1 ounce of milk

That means Nell's mom add 15 ml of chocolate syrup for each ounce of milk.

It is given that Nell's dad adds 65mL of chocolate syrup for every 5 ounces of milk.

This can be written as:

65 ml = 5 ounce of milk

Divide both the sides by 5.

13 ml = 1 ounce of milk

That means Nell's dad add 13 ml of chocolate syrup for each ounce of milk.

Since, Nell's mom made 15 ml of chocolate milk and his dad made 13 ml of chocolate milk, thus Nell's mom made chocolate milk more chocolaty.

Hence, option A is correct.

Dominik [7]3 years ago

5 0

Answer:

<em>Nell's mom's chocolate milk is more chocolatey.</em>

Step-by-step explanation:

Nell's mom makes chocolate milk with 30mL of chocolate syrup for every 2 ounces of milk.

30/2 = 15ml

<em>It means 15 ml for 1 ounce of milk</em>

Nell’s dad adds 65 ml of chocolate syrup for every 5 ounces of milk.

65/5 = 13 ml

<em>It means 13 ml for 1 ounce of milk</em>

Thus 15 ml > 13 ml

This shows that:

<u>Nell's mom's chocolate milk is more chocolatey</u>

<u></u>

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<h3>Answer: Choice D </h3>

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Explanation:

Let's go through the answer choices one by one to see which are true, and which are false.

Choice A) This is true because as we approach x = 2 from the left hand side, the y values get closer to y = 1 from the top
Choice B) This is true. As we get closer to x = 4 on the left side, the blue curve is heading downward forever toward negative infinity. So this is what y is approaching when x approaches 4 from the left side.
Choice C) This is true also. The function is continuous at x = -3 due to no gaps or holes at this location, so that means its limit here is equal to the function value.
Choice D) This is false. The limit does exist and we find it by approaching x = -4 from either side, and we'll find that the y values are approaching y = -2. In contrast, the limit at x = 2 does not exist because we approach two different y values when we approach x = 2 from the left and right sides (approach x = 2 from the left and you get closer to y = 1; approach x = 2 from the right and you get closer to y = -2). So again, the limit does exist at x = -4; however, the function is not continuous here because its limiting value differs from its function value.
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Step-by-step explanation:

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klio [65]

The smallest amount of material needed is 54 square centimeters

<h3>How to determine the amount of material needed?</h3>

The given parameters are:

Volume = 36 cubic centimeters

Represent length with x, width with y and height with z.

So, we have

x = 3y

The volume is calculated as:

V = xyz

This gives

V = 3y²z

Substitute 36 for V

3y²z = 36

Divide by 3

y²z = 12

Make z the subject

z = 12/y²

The surface area is:

S = 2(xy + xz + yz)

This gives

S = 2(3y² + 3yz + yz)

Evaluate the like terms

S = 2(3y² + 4yz)

Expand

S = 6y² + 8yz

Substitute z = 12/y²

S = 6y² + 8y * 12/y²

This gives

S = 6y² + 96/y

Differentiate

S' = 12y - 96/y²

Set to 0

12y - 96/y² = 0

Multiply through by y²

12y³ - 96 = 0

Add 96 to both sides

12y³ = 96

Divide by 12

y³ = 8

Take the cube root of both sides

y = 2

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x = 3y and z = 12/y²

This gives

x = 3 * 2 = 6

z = 12/2² = 3

Recall that:

S = 2(xy + xz + yz)

So, we have:

S = 2(6 * 2 + 3 * 3 + 2 * 3)

Evaluate

S = 54

Hence, the smallest amount of material needed is 54 square centimeters

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2 years ago

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denpristay [2]

Answer:

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that $\mu = 118, \sigma = 8$

Sample of 16 shells

This means that $n = 16, s = \frac{8}{\sqrt{16}} = 2$

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{120 - 118}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.841

X = 116

$Z = \frac{X - \mu}{s}$

$Z = \frac{116 - 118}{2}$

$Z = -1$

$Z = -1$ has a pvalue of 0.159

0.841 - 0.159 = 0.682

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

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