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vampirchik [111]
2 years ago
10

4. The data set DDT (MASS) contains independent measurements of the pesticide DDT on kale. Make a histogram and a boxplot of the

data. From these, estimate the mean and standard deviation. Check your answers with the appropriate functions.

Mathematics
1 answer:
padilas [110]2 years ago
8 0

Answer:

Step-by-step explanation:

Rcode:

library(MASS)

data(DDT,package="MASS")

head(DDT)

colnames(DDT)

print(DDT)

boxplot(DDT)

hist(DDT,main="Histogram for DDT")

fivenum(DDT)

dim(DDT)

outlier_values <- boxplot.stats(DDT)$out # outlier values.

boxplot(DDT, main="DDT", boxwex=0.1)

mtext(paste("Outliers: ", paste(outlier_values, collapse=", ")), cex=0.6)

mean(DDT)

sd(DDT)

Output: * see attachment below*

From Histogram the mean is at the peak

it lies in between 3and 3.5

Form Boxplot we found one outlier and it is 4.64

Fivenum summary is

2.790 3.075 3.220 3.360 4.640

min=2.790

Q1=3.075

Q2=3.220

Q3=3.360

max=4.640

From boxplot the center line is median which cuts the box into 2 equal halves

Distribution is symmetrical

Follows normal distribution.

Mean=3.328

Standard deviation=0.4371531

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Scilla [17]

Answer:  15 miles

<u>Step-by-step explanation:</u>

2/3 of the way is 30 miles

Let x represent the total distance to grandmother's house, then

\quad \dfrac{2}{3}x=30\\\\\rightarrow 2x=90\\\\\rightarrow x = 45


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3 0
2 years ago
List the potential solutions to 2 ln x = 4 ln 2 from least to greatest.
svp [43]

We have to find the potential solutions to 2 ln x = 4 ln 2 from least to greatest.

Using the properties of ln function.

a \times ln b = ln b^{a}

Therefore, we get

ln x^{2} = ln 2^{4}

ln x^{2} = ln 16

taking antilog on both the sides, we get

x^{2} = 16

So, x = \pm 4

Therefore, the potential solutions to 2 ln x = 4 ln 2 from least to greatest is -4 and 4.

3 0
2 years ago
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Greeley [361]

Answer:

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Step-by-step explanation:

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2 years ago
Use lagrange multipliers to find the points on the given surface that are closest to the origin. y2 = 64 + xz
Inessa [10]
The distance between an arbitrary point on the surface and the origin is

d(x,y,z)=\sqrt{x^2+y^2+z^2}

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D(x,y,z)=x^2+y^2+z^2

The Lagrangian would then be

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)

We have partial derivatives

\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}

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\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0

This means y^2=64\implies y=\pm8

so that the points on the surface closest to the origin are (0,\pm8,0).
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2 years ago
For each hour he babysits, Anderson earns $1 more than half of Carey’s hourly rate. Anderson earns $6 per hour. Which equation c
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Answer:

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