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andreyandreev [35.5K]

3 years ago

8

The table shows the current grades in one of the math classes and one of the science classes at a local high school. Find P(grad

e of B | science class).

2 answers:

Alenkinab [10]3 years ago

8 0

Hey, did you ever figure it out?

max2010maxim [7]3 years ago

4 0

Answer:

.667

Step-by-step explanation: 8/12 i may be wrong but 0.154 is def wrong

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Confused on this help plss

ValentinkaMS [17]

Answer:

of iv

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3

x+2({\color{#c92786}{4-x}})=2x+6-3x

x+2(4−x)=2x+6−3x

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2

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Step-by-step explanation:

follow me

8 0

3 years ago

James and Cameron are watching a film. They pause it with one fifth left to watch. What is the ratio of how much they have watch

goldenfox [79]

Answer:

The ratio of how much they have watched to how much they have left to see is 4 : 1.

Step-by-step explanation:

Consider that the total duration of the film is, 1.

It is provided that James and Cameron paused when there is one-fifth left to watch.

That is, they have already watched, $1-\frac{1}{5}=\frac{4}{5}$ th of the film.

Compute the ratio of how much they have watched to how much they have left to see as follows:

$\frac{\text{Duration Watched}}{\text{Duration Left}}=\frac{4/5}{1/5}=\frac{4}{1}=4:1$

Thus, the ratio of how much they have watched to how much they have left to see is 4 : 1.

5 0

3 years ago

Hello fellow Brainliers. I was wondering if anyone out there would like to go to school for me during the next school year and a

deff fn [24]

Answer:

I would not like to take your place but I can help you find someone. I don't have a laughing emoji. (On my computer)

6 0

3 years ago

Correct answer gets brainlest

Leona [35]

Answer:

1. 19(25x2)

Step-by-step explanation:

19 x 25 = 475

475 x 2 = 950

4 0

3 years ago

Read 2 more answers

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