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3 years ago

Does any one else have to do edgenuity

Mathematics

2 answers:

omeli [17]3 years ago

8 0

Answer:

me i hate it but it not that hard tho

Step-by-step explanation:

Sholpan [36]3 years ago

6 0

Yup but you can find almost all the answers online

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Fill the missing values in the ratio table

kramer

Answer:

A really out of focus picture, but I think this might be what you want.

Step-by-step explanation:

8 0

3 years ago

Please answer 7/10k=21 find x ill give you brainliest

algol [13]

Answer:

k=2.1

Step-by-step explanation:

7/10k=21

x7 x7 Multiply seven from each side of the equation.

10k=21

/10 /10 Divide ten from each side of the equation. 21/10=2.1

k=2.1

3 0

3 years ago

A rectangular magazine cover is photocopied by using a scale factor of One-third. The length of the photocopy of the magazine co

seraphim [82]

Answer: "reduced by a factor of One-third."

Step-by-step explanation:

Suppose that the original magazine has a length L and a width W.

If it is photocopied using a scale factor of K, then the measures of the photocopy will be:

length = K*L

Width = K*W

In this case, the scale factor is K = 1/3, then the measures of the photocopy will be:

length = (1/3)*L = L/3

Width = (1/3)*W = W/3

For usual notation:

When k > 1, we have an enlargement by a factor k

when 0 < k < 1, we have a reduction by a factor k

in this case, k = 1/3, then:

We have a reduction by a factor of 1/3

The correct option is:

"reduced by a factor of One-third."

7 0

3 years ago

Read 2 more answers

Solve. 489.209 + 43.9

lidiya [134]

Answer:

533.109

Step-by-step explanation:

Just add all the numbers

8 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

4 years ago

Read 2 more answers