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gizmo_the_mogwai [7]

3 years ago

8

Television sizes are described by the length of their diagonal measure. What would be the listed size of the TV shown? (to the n

earest whole inch)
1,424
32
38
20

1 answer:

prisoha [69]3 years ago

3 0

Answer:

38 inches

Step-by-step explanation:

since we are finding the diagonal we are going to use the equation a^2+b^2=c^2

so if we do substitution we would have 32^2+20^2=C^2

32^2= 1024

20^2= 400

so,

1024 + 400 = C^2

1424 = C^2

to get rid of the C^2 we have to take the square root of it. But whatever we do on one side of the equation we have to do on the other side so,

C = sqrt(1424)

C = 37.73

C = 38 inches

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6 0

3 years ago

One of the legs of a right triangle measures 9cm and the other leg measures 8cm. Find the measure of the hypotenuse. If necessar

Aleks04 [339]

Answer:

12.0 cm

Step-by-step explanation:

Use Pythagorean theorem $a^2+b^2 = c^2$

a = 9

b = 8

hypotenuse = c = $\sqrt{8^2+9^2}$

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4 0

2 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

I need help fastt plss

valina [46]

Answer:

102? hopefully this helps

7 0

3 years ago

A rectangular carpet has a perimeter of 264 inches. The length of the carpet is 108 inches more than the width. Determine the di

azamat

Answer:

The width of the carpet is 12 inches and the length of the carpet is 120 inches.

Step-by-step explanation:

$2w+2(w+108)=264$

^We can distribute first^

Distributing would result in:

$2w+2w+216=264$

Then we would combine like terms, which would result in:

$4w + 216 = 264$

Then we solve the equation.

$4w + 216 = 264 \\ \: \: \: \: \: \: \: \: \: \frac{ - 216 = - 216}{4w = 48} \\ \\ \frac{4w}{4} = \frac{48}{4} \\ w = 12$

Since the width of the carpet is 12 (as we figured out). The statement said that "The length of the carpet is 108 inches more than the width."

So now we add 108 to 12 to figure out the length.

$108 + 12 = 120$

The length of the carpet is 120 inches.

4 0

3 years ago