All categories

3 years ago

If $1,000 is invested at an

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

4 0

Answer:

probably 5 or so?

Step-by-step explanation:

I'm really not to sure myself but its around the 5-10s but if you're rounding at the nearest 430s it would be around the 20-30s or so if you ignore the order of operations just from just by doing mental maths I can conclude this with certainty.

You might be interested in

2. How many ounces of yeast will you need to make 30 rolls?

kupik [55]

Answer:

Step-by-step explanation:

7 0

3 years ago

a waiter had nine tables he was waiting on, with seven women and three men at each table. how many customees total did the waite

Greeley [361]

Since each table had 7 women and 3 men at each table, that would mean each table has 10 people. Since there are 9 tables he was waiting on, and 10 people at each table, multiply 9 x 10, and you will find that he served 90 people in total.

8 0

3 years ago

Read 2 more answers

What is the volume of the cylinder 7ft 18ft

Mademuasel [1]

Answer:

V=πr2h

Step-by-step explanation:

5 0

3 years ago

At a yearly basketball tournament, 64 different teams compete. After each round of the tournament, one-fourth of the teams remai

artcher [175]

Answer:

64*(1/4)^x

Step-by-step explanation:

64 is the starting number because it is given, and x represents how many rounds there are. 1/4 represents how much of the teams remain after each round.

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