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ioda
3 years ago
11

Jessie estimates the weight of her cat to be 10 pounds. The actual weight of the cat is 13.75 pounds. find the percent error.

Mathematics
1 answer:
romanna [79]3 years ago
3 0

Answer:

\large \boxed{27 \, \%} }

Step-by-step explanation:

\begin{array}{rcl}\text{Percent error}&= &\dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \,\%\\\\& = & \dfrac{\lvert 10 - 13.75\lvert}{13.75} \times 100 \, \% \\\\& = & \dfrac{\lvert -3.75\lvert}{13.75} \times 100 \, \%\\ \\& = & \dfrac{3.75}{13.75} \times 100 \,\%\\\\& = & 0.27 \times 100 \, \%\\& = & \mathbf{27 \, \%}\\\end{array}\\\text{The percent error is $\large \boxed{\mathbf{27 \, \%} }$}

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Step-by-step explanation:

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Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your
Veronika [31]

The expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Given an integral \int\limits^5_b {1} \, x/(2+x^{3}) dx.

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty}∑f(a+iΔx)Δx ,here Δx=(b-a)/n

\int\limits^5_1 {x/(2+x^{3}) } \, dx=f(x)=x/2+x^{3}

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)=[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}

\lim_{n \to \infty}∑f(a+iΔx)Δx=

\lim_{n \to \infty}∑n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n

=4\lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3}

Hence the expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

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