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3 years ago

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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3 years ago

In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion

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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> ( $\mu_1-\mu_2$ ) =

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

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