Round 1149 to the nearest thousand.

beks73 [17]

Answer:

C

Step-by-step explanation:

Mean is just another name for average. To find the mean of a data set, add all the values together and divide by the number of values in the set. The result is your mean!

5 0

3 years ago

Read 2 more answers

What is 1.83 repeated as a fraction

Oksi-84 [34.3K]

As a fraction, 1.83 is 1 83/100

7 0

3 years ago

Read 2 more answers

Write an equation in slope-intercept form of the line that passes through (-1, 4) and (0,2).<br> y =

telo118 [61]

Answer:

y =( -1/2 )x + 2

Step-by-step explanation:

first step is to determine the slope of the line ( which is the rise over the run) or symbolically slope is defined as m= ∆x / ∆y, so plugging those values we get...

m= ∆x / ∆y = (-1 - 0) / (4 - 2) = -1 / 2

so next is to find the zero( y-intercept) of the function by ....

y = mx + b

y = ( -1/2)x + b (since m is equal to -1/2)

2 = ( -1/2)0 + b

2= b

3 0

3 years ago

Which answer is a reasonable estimate to the following problem? 94 · 689

tino4ka555 [31]

94x689=64,766
if you round it to the nearest ten you get 90x690=62100
62100 is closer to 60,000 than it is to 70,000.
so you would round it down to 60,000.

8 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.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$