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3 years ago

What is the value of x?

Mathematics

1 answer:

nikklg [1K]3 years ago

4 0

It took me a minute, but I understand what it wants you to do.

a^2+b^2=c^2

Find the missing length on that one triangle to the left.

62^2+b^2= 99.2^2

Square the numbers.

3844+b^2= 9840.64

Subtract 3844 on both sides.

b^2= 5996.64

Square root it.

b= 77. 437975 round down to b= 77.4

Now, we can find the length of the other missing side. (x+2)

77.4^2+b^2= 112^2

5990.76+b^2= 12,544

Subtract 5990.76 on both sides.

b^2= 6553.24

Square root it.

b= 80.952085, round to b=81

Now, plug that in.

(x+2)= 81

Subtract 2 on both sides.

x=79 (about)

I hope this helps! Make sure to check my work.
~kaikers

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Evaluate 5a+3 4b when a = b - 4and b = -2.

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⊙O and ⊙P are given with centers (−2, 7) and (12, −1) and radii of lengths 5 and 12, respectively. Using similarity transformati

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Answer:

Whereby circle $\bigodot$ P can be obtained from circle $\bigodot$ O by applying the transformations of a translation of T₍₁₄, ₋₈₎ followed by a dilation by a scale factor of 2.4, $\bigodot$ O is similar to $\bigodot$ P

Step-by-step explanation:

The given center of the circle $\bigodot$ O = (-2, 7)

The radius of $\bigodot$ O, r₁ = 5

The given center of the circle $\bigodot$ P = (12, -1)

The radius of $\bigodot$ P, r₂ = 12

The similarity transformation to prove that $\bigodot$ O and $\bigodot$ P are similar are;

a) Move circle $\bigodot$ O 14 units to the right and 8 units down to the point (12, -1)

b) Apply a scale of S.F. = r₂/r₁ = 12/5 = 2.4

Therefore, the radius of circle $\bigodot$ O is increased by 2.4

We then obtain $\bigodot$ O' with center at (12, -1) and radius r₃ = 2.4×5 = 12 which has the same center and radius as circle $\bigodot$ P

∴ Circle $\bigodot$ P can be obtained from circle $\bigodot$ O by applying similarity transformation of translation of T₍₁₄, ₋₈₎ followed by a dilation by a scale factor of 2.4, $\bigodot$ O is similar to $\bigodot$ P.

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A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

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Answer:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=12.8$

The sample deviation calculated $s=4.324 \approx 4.3$

$12.8-4.604\frac{4.324}{\sqrt{5}}=3.896$

$12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704$

So on this case the 99% confidence interval would be given by (3.896;21.704)

Step-by-step explanation:

Data: 10,9,20,13,12

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=12.8$

The sample deviation calculated $s=4.324$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=5-1=4$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,4)".And we see that $t_{\alpha/2}=4.604$

Now we have everything in order to replace into formula (1):

$12.8-4.604\frac{4.324}{\sqrt{5}}=3.896$

$12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704$

So on this case the 99% confidence interval would be given by (3.896;21.704)

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