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sergey [27]
3 years ago
14

Nut Mix Michael and Kevin return to the candy store, but this time they want to purchase nuts. They can't decide among peanuts,

cashews, or almonds. They again agree to create a mix. They bought 2.5 pounds of peanuts for $1.30 per pound, 4 pounds of cashews for $4.50 per pound, and 2 pounds of almonds for $3.75 per pound. Determine the price per pound of the mix.
Mathematics
1 answer:
Harrizon [31]3 years ago
8 0
2.5x1.30=3.25
4x4.50=18
2x3.75=7.5
3.25+18+7.5=$28.75 total price
2.5+4+2=8.5lb. total weight
28.75/8.5= 3.38 dollars per pound.
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I need help!!
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Answer:

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Step-by-step explanation:

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A system of linear equations is graphed. Which ordered pair is the best estimate for the solution to the system? (12, 0) (1, 1)
GuDViN [60]
Line 1--------- >(0,2)     (1,-1)
y=mx+b         m=(-1-2)/(1-0)=-3
2=(-3)*(0)+b----- > b=2
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Line 2--------- >(0,-1)     (2,2)
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<span>using a graphic tool
see attached figure</span>

the best estimate solution for the system is <span>(1, 1) 
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3 years ago
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(-20, -16), (7, 20)<br> Find the slope of the line through<br> each pair of points.
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4 years ago
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There are 10 employees in a particular division of a company. Their salaries have a mean of $70,000, a median of $55,000, and a
alekssr [168]

Answer:

A) bar{x}_{new}=160,000

B) Median remains the same.

C) sigma_{new}=300998.34

Step-by-step explanation:

Consider the complete question attached below.

No. of employees = n = 10

Given mean = $70,000

Median = $55,000

Standard deviation = $60,000

Largest number on the list = $100,000

Accidentally changed to = $1,000,000

Modified mean, media, SD =?

A) Modified Mean:

\bar{x}=\frac{\sum x}{n} = 70,000\\\\\sum x=(\bar{x})(n) = (70,000)(10)\\\\\sum x=700,000\\\\\sum x_{new} =700,000 -100,000+1,000,000\\\\\sum x_{new}=1,600,000\\\\\bar{x}_{new}=\frac{1,600,000}{10}\\\\\bar{x}_{new}=160,000

B) Modified Median:

Median remains same and is not affected by changing highest value.

C) Modified SD:

Standard deviation is given by formula:

\sigma=\sqrt{\frac{\sum x^{2}-n\bar{x}}{N-1}}---(1)\\\\\sigma_{new}=\sqrt{\frac{\sum x_{new}^{2}-n\bar{x}_{new}}{N-1}}---(2)\\\\From\,\, (1)\\\\\sum x^{2}=(N-1)\sigma^{2}+n\bar{x}

\sum x^{2}=(10-1)(60,000)^{2}+(10)(70,000)^{2}\\\\\sum x^{2}=8.14\times 10^{10}\\\\\sum x_{new}^{2}= 8.14\times 10^{10}-(10,0000)^{2}+(1,000,000)^{2}\\\\\sum x_{new}^{2}=1.0714\times 10^{12}\\\\Using\,\, (2)\\\sigma_{new}=\sqrt{\frac{1}{9}(1.0714\times 10^{12}-(10)(1.6\times 10^{5})}\\\\\sigma_{new}=\sqrt{9.06\times 10^{11}}\\\\\sigma_{new}=300998.34

7 0
3 years ago
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