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2 years ago

Please help on test

Mathematics

1 answer:

dolphi86 [110]2 years ago

3 0

150 * 5/100=7.5

150+7.5=157.5$

145*7/100= 10.15

145+10.15= 155.15$

140*8/100= 11.2

140+11.2= 151.2$

135*9/100= 12.15

135+12.25= 147.15$

Purchase D is least Expensive. 147.15$

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f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans

Kruka [31]

Answer:

$\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558$

Step-by-step explanation:

We want to find the Riemann sum for $\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx$ with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)$

where $\Delta{x}=\frac{b-a}{n}$ .

Step 1: Find $\Delta{x}$

We have that $a=0, b=\frac{3\pi }{4}, n=6$

Therefore, $\Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}$

Step 2: Divide the interval $\left[0,\frac{3 \pi}{4}\right]$ into n = 6 sub-intervals of length $\Delta{x}=\frac{\pi}{8}$

$a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b$

Step 3: Evaluate the function at the left endpoints

$f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3$

$f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386$

$f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964$

$f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527$

$f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0$

$f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527$

Step 4: Apply the Left Riemann Sum formula

$\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315$

$\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558$

5 0

2 years ago

From 42 acres to 72 acres (round to the nearest percent

guajiro [1.7K]

Answer:

42 acres to 72 acres, there is an increase of 30 acres

percentage wise

72-42=30

30/72=.4166...

therefore there is also a 41.6% increase

Step-by-step explanation:

71.42 Percentage change:×100%

Now, the change is given from 42 acres to 72 acres, therefore

Change= 72-42=30

⇒Percentage change= ×100%

= 71.42

8 0

3 years ago

Andrea is a teacher. Her annual salary is $59,605. Her salary is dispersed to her semimonthly from September – June (10 months).

Brilliant_brown [7]

Answer:

don't know trying to get the ANSWER

7 0

3 years ago

A random sample of 20 houses selected from a city showed that the mean size of these houses is 1880 square feet with a standard

azamat

Answer:

Option C

Step-by-step explanation:

Given that for a random sample of 20 houses selected from a city showed that the mean size of these houses is 1880 square feet with a standard deviation of 320 square feet.

X, the sizes of houses is Normal

Hence sample mean will be normal with

Mean = 1880

and std dev = $\frac{320}{\sqrt{20} } \\=71.5542$

For 90% confidence interval critical value for t with degree of freedom 19 is

1.328

Confidence interval upper bound

= mean + margin of error

= $1880+1.328*71.55\\=1975$

Option C is right

3 0

3 years ago

The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is

bulgar [2K]

The integral is approximated by the sum,

$\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))$

where $f(x)=x^3$ and $x_n=\dfrac12n$ , giving you

$\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)$
$\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)$
$\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)$

Faulhaber's formulas make short work of computing the sum. You have

$\displaystyle\sum_{n=0}^k1=k+1$
$\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2$
$\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6$
$\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4$

which gives

$\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31$
$\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}$
$\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25$

4 0

3 years ago