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3 years ago

If R is the midpoint of QS, QR= 8x-51 and RS = 3x-6, find QS.

Mathematics

2 answers:

sashaice [31]3 years ago

7 0

QS would be 42. So the way i did it was I:

8x-51=3x-6

x=9

Plugged into both equations of QR and RS to then get the answer of 42

Mnenie [13.5K]3 years ago

5 0

Answer: The required length of QS is 42 units.

Step-by-step explanation: Given that the point R is the midpoint of the line segment QS,

where QR= 8x-51 and RS = 3x-6.

We are to find the length of QS.

Since R is the midpoint of the segment QS, so we must have

$QR=RS\\\\\Rightarrow 8x-51=3x-6\\\\\Rightarrow 8x-3x=51-6\\\\\Rightarrow 5x=45\\\\\Rightarrow x=\dfrac{45}{5}\\\\\Rightarrow x=9.$

Therefore, the length of QS is given by

$QS\\\\=QR+RS\\\\=8x-51+3x-6\\\\=11x-57\\\\=11\times9-57\\\\=99-57\\\\=42.$

Thus, the required length of QS is 42 units.

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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

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X when Z has a pvalue of 0.15. So X when Z = -1.047.

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3 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

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The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

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