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Effectus [21]
3 years ago
11

Keith wants to make cupcakes. To make cupcakes he needs 1 3/10 cups of flour per batch of cupcakes. If Keith has 13 cups of flou

r,then how many batches of cupcakes can Keith make?
Mathematics
1 answer:
marin [14]3 years ago
4 0

Answer: Keith can make 10 batches of cupcakes.

Step-by-step explanation:

To make cupcakes, he needs 1 3/10 cups of flour per batch of cupcakes. Converting 1 3/10 cups of flour into improper fraction, it becomes 13/10 cups of flour.

Let x represent the number of batches of cupcakes that Keith can make If Keith has 13 cups of flour, . Therefore,

13/10 cups of flour = 1 batch of cupcakes

13 cups of flour = x batch of cupcakes

Cross multiplying, it becomes

13/10 × x = 13 × 1

13x/10 = 13

13x = 13 × 10 = 130

x = 130/13 = 10

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Start with the number 2380.<br> Divide by 10,<br> The 8 will end up in the _____ place.
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The 8 will end up in the ones place.

Step-by-step explanation:

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3 years ago
Please help! Thank you so much
kykrilka [37]
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Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28, and 25. Compute the range, interquartile range, variance, and
Valentin [98]

So we have a set of integers,

X=\{15,20,25,25,27,28,30,34\}

And we need to compute some stats.

First the range,

The range of data set is difference between the maximum and the minimum of the data set.

So the minimum is 15 and maximum is 34 therefore the range is 34 - 15 = 19.

Second the interquartile range,

The interquartile range of data set is the difference of the first and third quartiles.

First quartile is the value separating the lower quarter and higher three - quarters of the set. The first quartile is computed by taking median of the lower half of the sorted set. This is 15, 20, 25, 25 and it's median is 22.5

Third quartile is therefore the median of 27, 28, 30, 34 which is 29.

Next up, variance.

The variance of set measures how much data is spread out. For data set x_1,\dots, x_n with an average a,

\mathsf{var}(X)=\Sigma_{i=1}^{n}\dfrac{(x_i-a)^2}{n-1}

So first compute the average value which is actually a mean,

a=\dfrac{1}{2}\Sigma_{i=1}^{n}a_i

The sum \Sigma_{i=1}^{n}a_i of numbers in set is 204.

Divide this by number of elements in the set (8).

a=\dfrac{204}{8}=25.5

Then compute the variance,

\mathsf{var}(X)=\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}\approx\boxed{34.57}

And finally standard deviation,

Since have computed variance the standard deviation is nothing too hard. It's defined as a square root of variance,

\mathsf{sde}(X)=\sqrt{\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}}=\sqrt{34.57}\approx\boxed{5.88}

Hope this helps.

r3t40

6 0
3 years ago
The answer to 1/6+3/4
Harman [31]

Answer:

\frac{11}{12}

Step-by-step explanation:

Least Common multiplier of 6 and 4 : 12

Adjust fractions based on the LCM:

\frac{2}{12}+\frac{9}{12}=\frac{11}{12}

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Is it easy or difficult to keep all of your expenses and savings within the specified amount?
REY [17]
I would have to say easy
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