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Sholpan [36]
4 years ago
14

A CAT scan produces equally spaced cross-sectional views of a human organ that provide information about the organ otherwise obt

ained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 18, 59, 78, 93, 105, 118, 128, 63, 38, and 0. Use the Midpoint Rule with n = 5 to estimate the volume V of the liver.

Mathematics
1 answer:
Semmy [17]4 years ago
6 0

Answer:

V=1101cm^3

Step-by-step explanation:

You are given this data:

\left[\begin{array}{cccccccccccc}long&0&1.5&3&4.5&6&7.5&9&10.5&12&13.5&15&&\\Area&0&18&59&78&93&105&118&128&63&38&0\end{array}\right]

First, calculate the x points by dividing the total length in 5:

\Delta{x}=\frac{l_f-l_0}{5}= \frac{15-0}{5}= 3

x=3,6,9,12,15

Now you calculate the half point of the x axis intervals you just calculated:

x_h=1.5,4.5,7.5,10.5,13.5

and find the function values of each of them (the Area for each cut):

A(1.5) = 18

A(4.5)=78

A(7.5)=105

A(10.5)=128

A(13.5)=38

Now you have formed the rectangles (see diagram below).

To calculate the volume, just use the next equation given by the midpoint rule:

V=\Delta{x}\sum_1^5{h_{rectangle}}\\V=3\sum(18, 78, 105, 128, 38)\\V=3(367)\\V=1101cm^3

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liraira [26]

A would be the correct answer because he is up in elevation so that makes the 1200 positive and he goes down 20.4 which is negative and is goes that far down!!

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3 years ago
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It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando
notka56 [123]

Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

Therefore, we have;

t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

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Step-by-step explanation:

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