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3 years ago

CAN SOMEONE PLEASEEEE ANSERR THISSSSSSSS THANK YOU SO VERY MUCH

Mathematics

2 answers:

Lana71 [14]3 years ago

7 0

Use Pythagorean theorem.

Let x be the side we are looking to determine:

$x = \sqrt{20^2-8^2}=\sqrt{336}\approx 18.3$

the missing measure is approximately 18.3 units

Lapatulllka [165]3 years ago

6 0

Pyth. theorem
$\sqrt{20 { }^{2} - 8 {}^{2} }$

=18.33
=18.3 (corr. to the nearest tenth)

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The larger of two numbers is one more than three times the smaller number.If the sum of the numbers is 49, find the numbers.

nirvana33 [79]

Answer:

larger number x = 37 and smaller number y = 12

Step-by-step explanation:

Let the larger number be x and the smaller number be y.

Sum of the numbers is $x + y = 49$ ....(1)

∴ $y = 49 - x$

Now, 3 times of the smaller number = 3y

According to question,

$x = 3y + 1 ......(2)$

Now, by substituting the value of y in (2)

$x = 3 (49 - x) + 1$

$x = 147 - 3x + 1$

$x + 3x = 148$

$4x = 148$

$x = 37$

Now,y = 49 - x and $x = 37$

$y = 49 - 37 = 12$

So the larger number x =37 and smaller number y = 12.

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3 years ago

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3 years ago

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goldfiish [28.3K]

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3 years ago

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Can someone please help me with my math corrections?

lara [203]

Answer:

Step-by-step explanation:

y = 2x + 3 {y = mx + b}

In Function A, rate of change is 2

Function B:

Rate of change = Change in y÷ change in x

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3 years ago

Suppose a computer engineer is interested in determining the average weight of a motherboard manufactured by a certain company.

irakobra [83]

Answer:

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

Step-by-step explanation:

How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error(width) as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this item, we have:

$M = 0.4, \sigma = 0.75$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 2.575*\frac{0.75}{\sqrt{n}}$

$0.4\sqrt{n} = 1.93125$

$\sqrt{n} = \frac{1.93125}{0.4}$

$\sqrt{n} = 4.828125$

$\sqrt{n}^{2} = (4.828125)^{2}$

$n = 23$

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error(width) as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this item, we have:

$M = 0.5, \sigma = 0.75$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.5 = 1.96*\frac{0.75}{\sqrt{n}}$

$0.5\sqrt{n} = 1.47$

$\sqrt{n} = \frac{1.47}{0.5}$

$\sqrt{n} = 2.94$

$\sqrt{n}^{2} = (2.94)^{2}$

$n \cong 9$

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

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3 years ago