MATHEMATICS HIGH SCHOOL

42 IS 21% OF WHAT NUMBER

Answers

Answer 1

Answer:

For this case we can make use of the following rule of three:

42 -----------------------> 21%

x -------------------------> 100%

In this case we have the following variable

x: unknown number whose 21% is equal to 42-

From the rule of three we clear x.

We have then:

Answer:

42 is 21% of 200

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Express each as a single power of 2...

2^a X 2^b X 2^c

2^a+b X 2^a-b

Answers

i) $2 ^(a)*2 ^(b)* 2^(c)$
We know that $x^(a) * x^(b) = x^(a+b)$
Therefore,
$2 ^(a)*2 ^(b)* 2^(c)$ = $2 ^(a+b+c)$
ii) $2 ^(a+b) * 2 ^(a-b)= 2^((a+b+a-b))$
$= 2^(2a)$

Find the area of a circle circumscribed about an equilateral triangle whose side is 18 inches long.a. 81
b. 108
c. 243

Answers

$The\ area\ of\ a\ circle:A_O=\pi r^2\ \ \ \ /r-a\ radius/\nThe\ length\ of\ a\ radius\ of\ a circle\ circumscribed\ about\ an\nequilateral\ triangle:r=(a\sqrt3)/(3)\ \ \ /a-a\ lenght\ of\ a\ side\ the\ triangle\n-------------------------------\nr=(18\sqrt3)/(3)=6\sqrt3\ (in)\n\nA_O=\pi\cdot\left(6\sqrt3\right)^2=\pi\cdot6^2\cdot\left(\sqrt3\right)^2=\pi\cdot36\cdot3=108\pi\ (in^2)\n\n\pi\approx3.14\n\ntherefore\n\nA_O\approx108\cdot3.14=339.12\ (in^2).....$

See attached for work.
$s=18\rightarrow r=6\sqrt3\rightarrow A_\odot=\pi(6\sqrt3)^2=\boxed{108\pi\ in^2}$

Write 2 litres 700 ml to the nearest litre

Answers

the nearest litre should be 3

How has the Pythagorean Theorem helped us through life?

Answers

it made calculating speed when a sphere rolls down the slope much more accurate than guessing

Answer:

It has made calculations much easier in Math.

The figure below shows a trapezoid, ABCD, having side AB parallel to side DC. The diagonals AC and BD intersect at point O.If the length of AO is three times the length of CO, the length of BO is

one-third the length of AC

one-third the length of AB

three times the length of DO

three times the length of DC

Answers

Answer:

Length of BO is three times the length of DO.

Step-by-step explanation:

In trapezoid ABCD, $CD\ ||\ AB$

In ΔCDO and ΔABO,

$m\angle CDO=m\angle ABO$ (Alternate interior angles)
$m\angle DCO=m\angle BAO$ (Alternate interior angles)

So, $\Delta CDO\sim \Delta ABO$ according to Angle-Angle similarity.

Therefore, the ratio of corresponding sides will be same.

$\Rightarrow (OD)/(OB)=(OC)/(OA)$

$\Rightarrow (OD)/(OB)=(OC)/(3\cdot OC)$

$\Rightarrow (OD)/(OB)=(1)/(3)$

$\Rightarrow OB=3\cdot OD$

BO is the same length as AO, and DO is the same length as CO. therefore BO is three times the length of DO.

Determine the exact value of the covariance expression Cov(2m,e m
). Compute the approximate value for Cov(2m,e
m
) using the simulation method. Compare your results between the exact and simulated values. b) [6 Marks] Compute the exact value of the integral η=∫
1
5

y
2
e
y
dy. Estimate the integral using the Monte Carlo (MC) integration method with a sample size of (n=1000). Determine the approximate percentage error (ϵ) between the exact value and the MC value. c) [8 Marks] Use the code to answer questions that follow: s 3336 <- function (N,×0,a,c,m){ pseudo <- rep(0,N) pseudo [1] <- <0 for (i in 2:(N+1)) pseudo[i] < (a∗ pseudo [i−1]+c)% pseudou <- pseudo/m return (pseudou) \} Explain the two pseudorandom number generation (PNG) methods, and identify the one used in the R code. Suppose (a=11,c=56,x
0

=13m=15) use the PNG to generate 30 pseudorandom numbers. Test the hypothesis that the generated numbers are uniformly distributed.

Answers

Answer:

Step-by-step explanation:

To determine the exact value of the covariance expression Cov(2m, em), we need more information about the variables involved. The covariance between two random variables, X and Y, is calculated as the expected value of the product of the differences between each variable and their respective means. Without the means or additional information, we cannot calculate the exact value of the covariance.

For the simulation method, we can generate random samples for 2m and em, calculate their covariance, and repeat the process multiple times to estimate an approximate value for Cov(2m, em). The simulated value will depend on the specific values generated for 2m and em in each iteration.

b) To compute the exact value of the integral η = ∫1^5 y^2 e^y dy, we can use integration techniques such as integration by parts or substitution. However, without further information or specific instructions, it is not possible to determine the exact value of this integral.

To estimate the integral using the Monte Carlo (MC) integration method, we can generate random points within the interval [1, 5] and evaluate the function y^2 e^y at those points. The estimate is then obtained by taking the average of these function values and multiplying it by the interval length (5 - 1). Using a sample size of n = 1000 means generating 1000 random points.

To calculate the approximate percentage error (ϵ) between the exact value and the MC value, you would need to know the exact value of the integral, which is not provided in the question.

c) The given code represents a pseudorandom number generation (PNG) method. It generates pseudorandom numbers using a linear congruential generator (LCG) algorithm. The LCG algorithm is a simple and widely used method for generating pseudorandom numbers based on a linear recurrence relation.

The LCG algorithm is defined by the recurrence relation:

X(n+1) = (a * X(n) + c) mod m

In the code, the values a = 11, c = 56, x0 = 13, and m = 15 are used as parameters for the LCG algorithm. It generates 30 pseudorandom numbers by iterating the recurrence relation.

To test the hypothesis that the generated numbers are uniformly distributed, you can perform a statistical test, such as the chi-square test or the Kolmogorov-Smirnov test. These tests compare the distribution of the generated numbers to a uniform distribution.

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