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Flauer [41]
4 years ago
6

When Marcus visited the corner store the first day he bought 3 sodas and 5 bag of chips for a total of $4.05. On the second day

he bought 2 sodas and 3 bags of chips for a total of $2.60. How much does each soda and bag of chips cost?
Mathematics
2 answers:
suter [353]4 years ago
7 0

Answer:

a soda cost .65 and chips cost 1.3

Step-by-step explanation:

i hope this helps

julia-pushkina [17]4 years ago
3 0

Answer:$4.05 + $2.60=

Step-by-step explanation:

One Marcus visit the store two times in a day you must add Both sums

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Answer:

The new mean is 5.

The new standard deviation is also 2.

Step-by-step explanation:

Let the sample space of hours be as follows, S = {x₁, x₂, x₃...xₙ}

The mean of this sample is 4. That is,\bar x=\frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}=4

The standard deviation of this sample is 2. That is, s=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}=2.

Now it is stated that each of the sample values was increased by 1 hour.

The new sample is: S = {x₁ + 1, x₂ + 1, x₃ + 1...xₙ + 1}

Compute the mean of this sample as follows:

\bar x_{N}=\frac{x_{1}+1+x_{2}+1+x_{3}+1+...+x_{n}+1}{n}\\=\frac{(x_{1}+x_{2}+x_{3}+...+x_{n})}{n}+\frac{(1+1+1+...n\ times)}{n}\\=\bar x+1\\=4+1\\=5

The new mean is 5.

Compute the standard deviation of this sample as follows:

s_{N}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{n-1}\sum ((x_{i}+1)-(\bar x+1))^{2}\\=\frac{1}{n-1}\sum (x_{i}+1-\bar x-1)^{2}\\=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=s

The new standard deviation is also 2.

5 0
4 years ago
Does anybody know how to do this? It’s due tomorrow! I need help really bad
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Answer:

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Step-by-step explanation:

Just ask someone

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