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2 years ago

7. Andre woke up at 7:18 a.m. and went to bed at 10:00 p.m. How long was Andre awake?

Mathematics

2 answers:

joja [24]2 years ago

4 0

Answer:

15 hours 42 min

sorry if im wrong

djverab [1.8K]2 years ago

3 0

He has had 9 hours and 18 minutes sleep

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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

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

where $m_i$ is the midpoint of the $i$ -th subinterval,

$m_i=\dfrac{\ell_i+r_i}2$

Then the integral $I$ is equal to the sum of the integrals of each interpolation over the corresponding $i$ -th subinterval.

$I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt$

It's easy to show that

$\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)$

so that the value of the overall integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}$

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3 years ago

A survey of 32 different gas stations in Texas found the average price of gasoline to be $3.19 a gallon with a sample standard d

Leni [432]

Answer:

The value of test statistic is -4.1247

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $3.26 a gallon

Sample mean, $\bar{x}$ = $3.19 a gallon

Sample size, n = 32

Sample standard deviation, σ = $0.096

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 3.26\text{ dollars a gallon}\\H_A: \mu < 3.26\text{ dollars a gallon}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{3.19 - 3.26}{\frac{0.096}{\sqrt{32}} } = -4.1247$

Thus, the value of test statistic is -4.1247

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3 years ago

Evaluate f (x) = 2 x for x = -4, 0, and 5.

aleksklad [387]

The function f(x) is defined in to different ways depending on the value of X. When x is < 5, use the expression 2x - 1 to evaluate the function When x is ≥ 5, use the expression x2 - 3 for example f(1) = 2(1) - 1 = 1 f(10) = (10)2 - 3 = 97

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3 years ago

Need help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Rudiy27

I think the equation is y= 3/5x-1 1/2

3 0

3 years ago

What is 1/3 of 18500.00

choli [55]

$\frac{1}{3}\ of\ 18500.00\\\\ = \frac{1}{3}*18500.00\\\\ =\frac{1}{3}*18500.00\\\\ =6166.67$

5 0

3 years ago