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swat32
3 years ago
8

Rick bought a total of 8 pounds of steak and chicken. If steak costs 13.50 per pound and chicken costs 3.25 per pound and he pai

d a total of 77.25, How many pounds of steak did he purchase?
Mathematics
1 answer:
ohaa [14]3 years ago
4 0

Rick bought 5 pounds of steak.

Step-by-step explanation:

Given,

Cost per pound of steak = 13.50

Cost per pound of chicken = 3.25

Total bought = 8 pounds

Total paid = 77.25

Let,

x represents the pounds of steak purchased

y represents the pounds of chicken purchased

According to given statement;

x+y=8    Eqn 1

13.50x+3.25y=77.25    Eqn 2

Multiplying Eqn 1 by 3.25

3.25(x+y=8)\\3.25x+3.25y=26\ \ \ Eqn\ 3

Subtracting Eqn 3 from Eqn 2

(13.50x+3.25y)-(3.25x+3.25y)=77.25-26\\13.50x+3.25y-3.25x-3.25y=51.25\\10.25x=51.25\\

Dividing both sides by 10.25

\frac{10.25x}{10.25}=\frac{51.25}{10.25}\\x=5

Rick bought 5 pounds of steak.

Keywords: linear equation, elimination method

Learn more about elimination method at:

  • brainly.com/question/13035995
  • brainly.com/question/13062539

#LearnwithBrainly

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4 years ago
A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playi
Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

<em>Let X = number of names recalled</em>

SO, X ~ N(\mu = 19.41,\sigma^{2} = 3.17^{2})

The z-score probability distribution is given by ;

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where, \mu = mean number of names recalled = 19.41

            \sigma = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

     P(X > 15) = P( \frac{X-\mu}{{\sigma} } } > \frac{15-19.41}{3.17}  } ) = P(Z > -1.39)

                                                      = P(Z < 1.39) = 0.9177  {using z table}

<em>Therefore, proportion of the students recalled more than 15 names is </em><em>91.77%.</em>

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3 years ago
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