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rewona [7]
3 years ago
12

a recipe calls 1 2/3 cups of brown sugar for the granola bars and 1 q/3 cups of brown sugar for the topping. Dara has 3 1-4 cups

of brown sugar. Does she have enough brown sugar to make granola bars and topping?
Mathematics
1 answer:
sp2606 [1]3 years ago
4 0

Answer:

yes, she does have enough

Step-by-step explanation:

3 1/4 - 1 1/3 - 1 2/3

=1/4

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Natasha_Volkova [10]
108 in^2

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9 * 12 = 108

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6 0
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Which size yogurt has the lowest unit price?<br> 6oz=0.89<br> 8oz=1.04<br> 10oz=1.69<br> 32oz=4.79
Savatey [412]
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5 0
3 years ago
The height of a triangle is 4 times its base. The area of the triangle is 288 square. Find the base and the height of the triang
Vanyuwa [196]
\bf \textit{area of a triangle}\\\\&#10;A=\cfrac{1}{2}bh\qquad &#10;\begin{cases}&#10;b=base\\&#10;h=height\\&#10;----------\\&#10;h=4b\leftarrow &#10;\begin{array}{llll}&#10;\textit{height is 4}\\&#10;\textit{times the base}&#10;\end{array}\\&#10;A=288&#10;\end{cases}\implies 288=\cfrac{1}{2}\cdot b\cdot 2b

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7 0
3 years ago
Cheryl collected data for her mathematics project. She noted that the data set was approximately normal.
nadya68 [22]

Answer:

X_(r) >> X_(n)

The mean for this case would increase since is defined as:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

The interquartile range would not change since the definition for the IQR is IQR =Q_3 -Q_1 and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

Step-by-step explanation:

For this case we assume that we have a random sample given X_(1), X_(2) ,..., X_(n) and for each observation X_i \sim N(\mu, \sigma) since the problem states that the data is approximately normal.

Let's assume that the largest value on this sample is X_(n) and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:

X_(r) >> X_(n)

The mean for this case would increase since is defined as:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

The interquartile range would not change since the definition for the IQR is IQR =Q_3 -Q_1 and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

6 0
3 years ago
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