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boyakko [2]
3 years ago
11

A store is having a sale on walnuts and chocolate chips. For 2 pounds of walnuts and 5 pounds of chocolate chips, the total cost

is 10.00. For 8 pounds of walnuts and 3 pounds of chocolate chips, the total cost is 23.00. Find the cost for each pound of walnuts and each pound of chocolat chips
Mathematics
2 answers:
oksian1 [2.3K]3 years ago
8 0
It will be $2.50 for each pound of walnuts, and it will be $1 for each pound of chocolate chips.

I know this hold true because when we multiply $2.50 by 2, we get $5, and add that to the other $5 from the chocolate chips and we will get $10. We can also multiply $2.50 by 8 and add $3 from chocolate chips to get $20 + $3, or $23.
Julli [10]3 years ago
7 0
1.  2w+5c=10
2.  8w+3c=23
Simplify the first equation: w=(10-5c)/2
Substitute: 8((10-5c)/2)+3c=23
Simplify:(80-40c)/2+3c=23
40-20c+3c=23,
17c=17, 1 pound of chocolate chips is $1
Substitute: 2w+5=10, 1 pound of walnuts is $2.50
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kondaur [170]

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Step-by-step explanation:

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4 0
2 years ago
Determine whether the statement describes a descriptive or inferential statistic. A survey of 3459 people revealed that 52% work
andriy [413]

Answer:

Inferential statistics.

Step-by-step explanation:

We have been given a statement. We are asked to determine whether the given statement describes a descriptive or inferential statistic.

Statement: A survey of 3459 people revealed that 52% work a full-time job; therefore it can be assumed that 52% of the U.S. population works a full-time job.

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7 0
3 years ago
If y-3x=9, 7x+y=25, what is the value of x and y?
lianna [129]

\bold{\rm{Given}}\text{ : If y - 3x = 9, 7x + y = 25, what is the value of x and y}?

\bold{\rm{Solution}} \text{: We'll solve this by substitution method}

\dashrightarrow  y - 3x = 9 \\  \\  \dashrightarrow   \boxed{y = 9 + 3x } \qquad .. \sf eq(1)

\text{we got the value of y now getting the value of x}

\looparrowright 7x + y = 25 \\  \\  \looparrowright 7x = 25 - y \\  \\  \looparrowright  \boxed{x =   \frac{25 - y}{7}} \qquad .. \sf eq(2)

\text{Now, putting y = 9 + 3x in eq(2)}

\hookrightarrow  x =  \frac{25 - y}{7}   \\  \\ \hookrightarrow  x =  \frac{25 - 9 + 3x}{7}  \\  \\ \hookrightarrow  7x =  25 - 9 + 3x \\  \\ \hookrightarrow  7x - 3x =  16 \\  \\ \hookrightarrow  4x = 16 \\  \\ \hookrightarrow  x =  \frac{16}{4}  \\  \\ \star \quad  \boxed{ \green{ \frak{ x = 4}}}

\text{Now we know the value of x, for getting y we need to put x in eq(1)}

: \implies y = 9 + 3x \\  \\   : \implies y = 9 + 3(4) \\  \\  : \implies y = 9 + 12  \\  \\    \star \quad  \boxed{\frak{  \green{y = 21}}}

\text{Hence, values of x and y are} \:  \frak{ \green{4}}  \: \text{and} \:  \frak{ \green{21}} \: \text{respectively}.

4 0
2 years ago
Two​ shooters, Rodney and​ Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark
Sedaia [141]

Complete Question

Answer:

a

  SE  = 0.66}

b

-3.29 <  \mu_1 - \mu_2 <  -0.70  

Step-by-step explanation:

From the question we are told that

  The sample size is  n  = 60

   The first sample mean is  \= x _1  =  8

    The second sample mean is   \= x _2  =  10

    The first variance is  v_1 =  0.25

    The first variance is  v_2 =  0.55

Given that  the confidence level is 95% then the level of significance is 5% =  0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the first standard deviation is  

     \sigma_1 =  \sqrt{v_1}

=>   \sigma_1 =  \sqrt{0.25}

=>   \sigma_1 =  0.5

Generally the second standard deviation is

     \sigma_2 =  \sqrt{v_2}

=>   \sigma_2 =  \sqrt{0.55}

=>   \sigma_2 =  0.742    

Generally the first standard error is

     SE_1  =  \frac{\sigma_1}{\sqrt{n} }

      SE_1  =  \frac{0.5}{\sqrt{60} }

     SE_1  =  0.06

Generally the second standard error is

     SE_2  =  \frac{\sigma_2}{\sqrt{n} }

      SE_2  =  \frac{0.742}{\sqrt{60} }

     SE_2  =  0.09

Generally the standard error of the difference between their mean scores is mathematically represented as    

      SE  =  \sqrt{SE_1^2 + SE_2^2 }

=>     SE  =  \sqrt{0.06^2 +0.09^2 }

=>     SE  = 0.66}

Generally 95% confidence interval is mathematically represented as  

      (\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } *  SE) <  \mu_1 - \mu_2 <  (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } *  SE)

=> (8 -10) -(1.96 *  0.66) <  \mu_1 - \mu_2 <  (8-10) +(Z_{\frac{\alpha }{2} } *  0.66)  

=>  -3.29 <  \mu_1 - \mu_2 <  -0.70  

 

5 0
3 years ago
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