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2 years ago

I’m confused to solve no.13 plz help me!!!

Mathematics

1 answer:

balu736 [363]2 years ago

6 0

Ask Google OK buddy you will get some nice help or go in some special maths videos

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13530 grams is equal to how many kilograms

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Answer: 13.53 kilograms (happy to help)

Step-by-step explanation:

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3 years ago

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Simplify the Equation Below Pls answer :(<br> 8x - 7 + 6x + 3

Liono4ka [1.6K]

Answer:

14x-4

Step-by-step explanation:

8x-7+6x+3

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2 years ago

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In a city school, 60% of students have blue eyes, 55% have dark hair, and 35% have blue eyes and dark hair. What is the probabil

gulaghasi [49]

Answer:

58%

Step-by-step explanation:

This is a problem of conditional probability.

Let A represent the event that student has dark hair.

So P(A) = 55% = 0.55

Let B represents the event that student has blue eyes.

So, P(B) = 60% = 0.60

Probability that student has blue eyes and dark hairs = P(A and B) = 35% = 0.35

We are to find the probability that a randomly selected student will have dark hair, given that the student has blue eyes. Using the given formula and values, we get:

$P(A|B)=\frac{P(A \cap B)}{P(B)}\\\\ P(A|B)=\frac{0.35}{0.60}\\\\ P(A|B)=0.58$

Therefore, there is 0.58 or 58% probability that the student will have dark hairs, given that the student has blue eyes.

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3 years ago

Consider the integral Integral from 0 to 1 e Superscript 6 x Baseline dx with nequals 25 . a. Find the trapezoid rule approximat

photoshop1234 [79]

Answer:

With n = 25, $\int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549$

With n = 50, $\int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594$

b. $\int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943$

The absolute error in the trapezoid rule is 0.08047

The absolute error in the Simpson's rule is 0.00008

Step-by-step explanation:

a. To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using n = 25 with the trapezoid rule you must:

The trapezoidal rule states that

$\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We have that a = 0, b = 1, n = 25.

Therefore,

$\Delta{x}=\frac{1-0}{25}=\frac{1}{25}$

We need to divide the interval [0,1] into n = 25 sub-intervals of length $\Delta{x}=\frac{1}{25}$ , with the following endpoints:

$a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

$2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579$

...

$2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701$

$f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the trapezoid rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549$

To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using n = 50 with the trapezoid rule you must:

We have that a = 0, b = 1, n = 50.

Therefore,

$\Delta{x}=\frac{1-0}{50}=\frac{1}{50}$

We need to divide the interval [0,1] into n = 50 sub-intervals of length $\Delta{x}=\frac{1}{50}$ , with the following endpoints:

$a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875$

$2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

...

$2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705$

$f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the trapezoid rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594$

b. To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using 2n with the Simpson's rule you must:

The Simpson's rule states that

$\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We have that a = 0, b = 1, n = 50

Therefore,

$\Delta{x}=\frac{1-0}{50}=\frac{1}{50}$

We need to divide the interval [0,1] into n = 50 sub-intervals of length $\Delta{x}=\frac{1}{50}$ , with the following endpoints:

$a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175$

$2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

...

$4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541$

$f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the Simpson's rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943$

c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by $|A-B|$

The absolute error in the trapezoid rule is

The calculated value is

$\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225$

and our estimate is 67.1519320308594

Thus, the absolute error is given by

$|67.0714655821225-67.1519320308594|=0.08047$

The absolute error in the Simpson's rule is

$|67.0714655821225-67.0715427161943|=0.00008$

6 0

2 years ago

the labelled price of a bag is Rs1,580. If 5% discount is allowed, calculate the selling price of the watch.solve it

Ipatiy [6.2K]

Step-by-step explanation:

If there 5% is allowed in Rs 1,580 than the selling price of the watch is 1,501

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3 years ago