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3 years ago

Hey can someone help me ? I’ll give brainliest !!

Mathematics

2 answers:

Alla [95]3 years ago

7 0

Answer:

132 kilometers

Step-by-step explanation:

Luda [366]3 years ago

4 0

7 bc I’m so smart thank you ciaoooooo

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Maria can read 20 pages of economics in an hour. She can also read 50 pages of sociology in a hour. She...? Maria can read 20 pa

denis-greek [22]

B.) Marua will spend 2 hours to read 100 pages of sociology which means she can only read 60 pages of economics.
Therefore, the opportunity cost of reading 100 pages of sociology is forfeiting 40 pages (2 hours) of economics.

8 0

3 years ago

Two variables are correlated with r = -0.23. Which description best describes the strength and direction of the association betw

tatuchka [14]

Answer is C weak negavite

weak, because as the value became smaller that 1 the correlation weakens. negavite because it is a negative value (-0.23)

3 0

3 years ago

Are 9/21 and 6/14 equivalent fractions

sesenic [268]

Answer:

Yes

Step-by-step explanation:

9/21 SIMPLIFIED= 3/7

6/14 SIMPLIFIED= 3/7

5 0

2 years ago

Read 2 more answers

An urn contains six red balls and four green balls. A sample of seven balls is selected at random.

KIM [24]

There are 10 balls in the Urn Total.

Red: 6

Green: 4

Question One: The probability that five red and two green is selected is likely. (as that is over half for both)

Question Two: Impossible. There is only 6 red balls, and 7 are taken from the urn. Thus it would at most be possible for 6 red and 1 green.

Question Three: At least four is likely, as there is more red then green in the Urn.

Hope I helped!

(Mark Brainliest if you can please!)

5 0

3 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago