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3 years ago

Equation for line parallel to 2x-3y+6=0, with the same y intercept

Mathematics

2 answers:

Makovka662 [10]3 years ago

8 0

I hope this helps you slope=2/3 x'= -3 y'= 2 slope (x'-x)=y'-y 2/3(-3-x)=2-y -6-2x=6-3y 3y-2x-12=0

e-lub [12.9K]3 years ago

3 0

Ok, so think for 1 second

the equation of a line that is paralell to the line with the SAME y intercept is the same line

so the answer is 2x-3y+6=0 or any other form of that like 2x-3y=-6, or y=2/3x+2

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An exit poll of 1000 randomly selected voters found that 515 favored measure A. a. Construct a 99% confidence interval for the s

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Answer:

Step-by-step explanation:

Given that an exit poll of 1000 randomly selected voters found that 515 favored measure A.

Sample proportion p = $\frac{515}{1000} =0.515$

$q=1-0.515 =0.485\\n =1000\\SE = \sqrt{\frac{pq}{n} } \\=0.158\\$

Margin of error 99% = 2.58*SE

= $2.58*0.0158\\=0.0408$

99% confidence interval = $(0.515-0.0408, 0.515+0.0408)\\= (0.474, 0.556)$

------------------------

$H_0: p =0.5\\H_a: p >0.5\\$

(Right tailed test)

STd error = $\sqrt{\frac{0.5*0.5}{1000} } \\=0.0158$

Test statistic Z = p diff/std error = $\frac{0.015}{0.0158} \\\\=0.9487$

p value = 0.1714

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3 years ago

Which of the following is equivalent to (16A^8B^4)^1/4 A)4A^4 B)4A^2B C)2A^2B D)2A^4

AlladinOne [14]

Answer:

Simplify.

Step-by-step explanation:

However is same thing

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3 years ago

Simplify (3x+1)^2(4x-2) hurry

Arisa [49]

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4 years ago

A company sells snack bags that contain a mixture of almonds and walnuts. There are 13 almonds per bag and 9 walnuts per bag. If

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There are 13 almonds and 9 walnuts per bag so that’s 13+9=22 nuts per bag.

If there are x bags then the number of nuts is 22x.

Assuming y is the total number of nuts we get y=22x

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4 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

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

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

99% confidence interval for ( $\mu_1-\mu_2$ ) =

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

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

= [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

4 years ago

Read 2 more answers