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3 years ago

Jose tosses a fair coin 2 times. What is the probability it will land on heads two times in a row?

Mathematics

2 answers:

sineoko [7]3 years ago

4 0

Answer:

1/2

Step-by-step explanation:

Umnica [9.8K]3 years ago

3 0

1/2 is the answer. hope this helps

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Jasmine baby-sits for $8.25 per hour. If she baby-sits for 12 hours in one weekend, how much money will she earn? Write your ans

forsale [732]

Answer:

Step-by-step explanation:

8.25*12= $99.00

7 0

2 years ago

Someone pls help me I'm stuck on this question!!!!????

Iteru [2.4K]

What’s the questions

5 0

3 years ago

Graph the function <br> Is it a, b, c or d?

Jobisdone [24]

Answer:

Step-by-step explanation:

(the line is a / shape, so the slope is positive)

(the line crosses the y axis at (0,-2))

(using the above information, we can conclude that the equation is y=2x-2)

slope = m = rise/run = 2/1 = 2

y=mx+b

y=2x+b

Choose any point on the line) (1,0)

0=2(1)+b

b=-2

y=2x-2

6 0

3 years ago

Who’s right? And why?

kvasek [131]

Answer: Raymond is correct

The DJ charges $75 an hour.

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Explanation:

x = number of hours

y = total cost

The DJ does a 3-hour event for $275. This means one point is (x,y) = (3,275)

Another point is (5,425) because a 5-hour event costs $425.

Let's find the slope of the line through these two points.

$(x_1,y_1) = (3,275) \text{ and } (x_2,y_2) = (5,425)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{425 - 275}{5 - 3}\\\\m = \frac{150}{2}\\\\m = 75\\\\$

The slope is 75, meaning that each time y goes up by 75, x goes up by 1.

Flipped around it means each time x goes up by 1, y goes up by 75.

This means each time we add on an extra hour, the cost goes up by $75.

The DJ charges $75 an hour. Raymond is correct.

3 0

2 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