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4 years ago

A meteorologist forecasted that Monday's high temperature would be 76°F. On Monday,

Mathematics

2 answers:

goldenfox [79]4 years ago

6 0

The percent error in the meteorologist's forecast for Monday is 5.26316 %

<h3><u>Solution:</u></h3>

Given that Monday's high temperature would be 76°F

On Monday, the temperature reached 80°F

To find: percent error in the meteorologist's forecast for Monday

Percent error is the difference between a measured and known value, divided by the known value, multiplied by 100%

<em><u>The percent error is given as:</u></em>

$\text {percent error }=\frac{\text {observed - standard}}{\text { standard value}} \times 100$

Here standard value = 76 and observed value on monday = 80

$\text { percent error }=\frac{80-76}{76} \times 100$

$\text { percent error }=\frac{4}{76} \times 100=5.26316 \%$

Thus percent error in the meteorologist's forecast for Monday is 5.26316 %

OlgaM077 [116]4 years ago

4 0

Answer:

Step-by-step explanation:

i got it right

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3 years ago

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If r and s are the 2 solutions of the equation below and r>s, what is the value of r-s?

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3 years ago

A spinner contains the numbers 1 through 50. What is the probability that the spinner will land on a number that is not a multip

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3 years ago

The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

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Answer:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Step-by-step explanation:

The probability (P) to find the particle is given by:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

The solution of the intregral of equation (1) is:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

I hope it helps you!

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4 years ago