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3 years ago

Is the product of A fraction less than 1 and a whole number greater than or less than the whole number?

1 answer:

3 0

It is less than the whole number.

Why? because 1*WholeNumber = WholeNumber. So if you multiply it by anything smaller than 1, than the result has to be smaller than the result you get by multiplying by one.

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The first 5 terms of a sequence are shown below:

nirvana33 [79]

Step-by-step explanation:

To find the formula for the nth term ,

First , find the common difference or common ratio.

Common difference =

$T_2-T_1 \\ - 2 - 3 \\ = - 5$

Common ratio =

$T_2/T_1 \\ \frac{ - 2}{3}$

The common difference applies to the other given numbers for example :

$T_3- T_2 = - 7 - ( - 2) \\ = - 7 + 2 = - 5$

this proves that it is an arithmetic sequence and the given formula for the nth term of an arithmetic sequence is

T_n =a(n-1)d

Where n = no of terms

a= first term

d= common difference

6 0

2 years ago

What shape is this?<br> A.cylinder<br> B.square<br> C.rectangle<br> D.tetrahedron

Mice21 [21]

Thats a rectangle .........

6 0

2 years ago

A particular fruit's weights are normally distributed, with a mean of 618 grams and a standard deviation of 24 grams. If you pic

Lorico [155]

Answer: The probability is 0.5136.

To find this probability we have to find the range of the values on the normal distribution table.

First, we need to find the percent for each making z-score.

Lower end: 50/24 = 2.08 = 0.0183

Upper end: 2/24 = 0.08 = 0.5319

Now, that we have the upper and lower ends, we can subtract them to find the difference or the range.

0.5319 - 0.0183 = 0.5136

There is a 51.36% chance they are in the given range.

3 0

2 years ago

In which way is a debit cat similar yo a check

katrin [286]

A debit card is similar to a check because they are both taking money out of a bank account under your name.

3 0

2 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

2 years ago