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Paladinen [302]
3 years ago
12

A pencil at a stationery store costs $1, and a pen costs $1.50. Stefan spent $7 at the store. He bought a total of 6 items. Whic

h system of equations can be used to find the number of pencils (x) and pens (y) he bought?
Mathematics
2 answers:
Elena-2011 [213]3 years ago
7 0
Let pencil represent x
pen represent y
Next, we have this two equation:
x+1.5y=7
x+y=6
Then, I will use substitution to find the value of x and y:
x+1.5y=7
-
x+y=6
=
0.5y=1
y=2 pens
x+y=6
x+2=6
Subtract 2 for both side
x+2-2=6-2
x=4 pencils
Check:
x+y=6 items
4+2=6 items Right

x+1.5y=7
4+1.5(2)=$7
$4+$3=$7
$7=$7
As a result, there are 4 pencils , and 2 pens. Hope it help!
Nina [5.8K]3 years ago
4 0
Y= # of pencils 
z= # of pens
1y + 1.50z = 7
Total # of pencils and pens combined equals 6 
4 pencils = $4
2 pens = $3 
$4 + $3 = $7 
4 pencils + 2 pens = 6 items total
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Can somone help me with number 23 and 24
tangare [24]

23.

store price = 14 + 0.5(14)

store price = 21

customer price = 21 + 0.08(21)

customer price = 21 + 1.68

customer price = $22.68

24.

formula: A = P(1 + rt)

A = 140(1 + (0.03)(2))

A = 140(1.06)

A = 148.40

interest = 148.40 = 140

interest = $8.40

Hope this helps!! :)

7 0
4 years ago
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A town doubles it’s size every 35 years if the population is currently 5,00, what will be the population be in 140 years?
Doss [256]

Answer:

8,000

Step-by-step explanation:

after 35 years 1000

after 70 2000

after 105 4000

qfter 140 8000

i might be wrong but hope this helped there is probs a faster way to do this tho.

5 0
3 years ago
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Yanka [14]

Answer:

C is the answer

Step-by-step explanation:

Please mark brainliest

7 0
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dsp73

Answer:

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Step-by-step explanation:

3 0
3 years ago
Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin
Artemon [7]
"<span>Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find \bar{d}, s_{d}, the t-test statistic, and the critical values to test the claim that \mu_{d} = 0"

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find </span><span>\bar{d}
You are asked to find the mean difference between the two variables, which is given by the formula:
\bar{d} =  \frac{\sum (x - y)}{n}

These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example: 
</span><span>\bar{d} =  \frac{6}{8} = 0.75</span>

B) Find <span>s_{d}
</span><span>You are asked to find the standard deviation, which is given by the formula:
</span>s_{d} =  \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }

These are the steps to follow:
1) Subtract the mean difference from each pair's difference 
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:
s_{d} = \sqrt{ \frac{101.5}{8-1} }
= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:
t =  \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}

where SE = standard error is given by the formula:
SE =  \frac{ s_{d} }{ \sqrt{n} }

These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract  \mu_{d}
6) divide by the standard error

In our example:
SE = 3.81 / √8
      = 1.346

The problem gives us <span>\mu_{d} = 0, therefore:
t = [(9.75 - 9) - 0] / 1.346</span>
  = 0.56

D) Find t_{\alpha / 2}
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find <span>t_{\alpha / 2} = 1.895</span>

Since our t-value is less than <span>t_{\alpha / 2}</span> we can reject our null hypothesis!!

7 0
4 years ago
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