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umka21 [38]
1 year ago
14

Help me solve this problem

Mathematics
1 answer:
Gennadij [26K]1 year ago
3 0

The value of GK = 18.6, GM = 14.5 AND JZ = 19.9.

From the figure:

MZ is a perpendicular bisector of ∆GHJ.

GM = MJ

GM = 14.5

KZ is a perpendicular bisector of ∆GHJ.

GK = KH

GK = 18.6

Z is the circumcenter of ∆GHJ.

JZ = GZ

JZ = 19.9

Therefore the value of GK = 18.6, GM = 14.5 AND JZ = 19.9.

Learn more about the perpendicular bisector here:

brainly.com/question/24753075

#SPJ1

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Ostrovityanka [42]
X+4+3x =
4x+4 (you sum the like terms )
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For the function given below, find a formula for the Riemann sum obtained by dividing the interval (0, 3) into n equal subinterv
Viktor [21]

Splitting up [0, 3] into n equally-spaced subintervals of length \Delta x=\frac{3-0}n = \frac3n gives the partition

\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]

where the right endpoint of the i-th subinterval is given by the sequence

r_i = \dfrac{3i}n

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Then the definite integral is given by the infinite Riemann sum

\displaystyle \int_0^3 2x^2 \, dx = \lim_{n\to\infty} \sum_{i=1}^n 2{r_i}^2 \Delta x \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac6n \sum_{i=1}^n \left(\frac{3i}n\right)^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3} \sum_{i=1}^n i^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \boxed{18}

8 0
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HELP!!!!<br> Please help me!!<br> I need help!!<br> HELP!!!!
mote1985 [20]

Answer:

Pretty sure it's A

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5 0
3 years ago
A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.
Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean(\mu) = 500 and a standard deviation(\sigma) =  100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = \dfrac{Y-\mu}{\sigma}

Z-score = \dfrac{625-500}{100}

⇒ Z-score = \dfrac{125}{100}

⇒Z-score =1.25

Therefore , the required z-score = 1.25

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