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VMariaS [17]
3 years ago
9

Emily Elizabeth has $1.50 in dimes and nickels. She has 22 coins in all. How many nickels does she have?

Mathematics
1 answer:
Marat540 [252]3 years ago
6 0

Answer:

14

Step-by-step explanation:

We can set up the following system of equations:

\begin{cases}10d+5n=150\\d+n=22\\\end{cases}, where d is the number of dimes she has and n is the number of nickels she has.

Multiplying the second equation by -5 and adding both equations lets us isolate and solve for d:

10d+5n+(-5(d+n))=150+(-5(22)),\\5d=40,\\d=8

Plugging d=8 into any of the two equations will let us solve for n:

8+n=22,\\n=\fbox{$14$}.

Therefore, she has 14 nickels.

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Answer:

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Step-by-step explanation:

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5 0
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twice the difference between a number and 10 is equal to 6 times the number plus 14. what is the number?
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Let the number be n.  Then 2(n-10) = 6n + 14.

Multiplying out the left side, 2n - 20 = 6n + 14.  Thus, -34=4n, and n = -17/2.
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3 years ago
Which of the following properties was used in the expression shown here?
ra1l [238]
Distributive property
5 0
3 years ago
In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion
S_A_V [24]

Answer:

99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Step-by-step explanation:

We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.

Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.

Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;

                              P.Q. = \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n__1+_n__2-2

where, \bar X_1 = average gold thickness of control-immersion tip plating = 1.5 μm

\bar X_2 = average gold thickness of total immersion plating = 1.0 μm

s_1 = sample standard deviation of control-immersion tip plating = 0.25 μm

s_2 = sample standard deviation of total immersion plating = 0.15 μm

n_1 = sample of printed circuit edge connectors plated with control-immersion tip plating = 7

n_2 = sample of connectors plated with total immersion plating = 5

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }   =  \sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2}  }{7+5-2} }  = 0.216

<em>Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

So, 99% confidence interval for the difference between the mean population mean, (\mu_1-\mu_2) is ;

P(-3.169 < t_1_0 < 3.169) = 0.99  {As the critical value of t at 10 degree of

                                              freedom are -3.169 & 3.169 with P = 0.5%}  

P(-3.169 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } < 3.169) = 0.99

P( -3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } ) = 0.99

P( (\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } ) = 0.99

<u>99% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } , (\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } ]

= [ (1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5}  } } , (1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5}  } } ]

= [0.099 μm , 0.901 μm]

Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

6 0
3 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

We are interested in the last two columns.

If we compare them we see that each value of the distance column is the absolute value of corresponding value of difference:

  • a - b        = 1, 5, - 3
  • distance  = 1, 5, 3

Corect choice is C

3 0
1 year ago
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