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3 years ago

What is an outlier in the data set 65, 32, 9, 76, 69?

1 answer:

6 0

Put the numbers in order
9,32,65,69,76
Q1 = (32 + 9) / 2 = 41/2 = 20.5
Q2 = 65
Q3 = (69 + 76) / 2 = 145/2 = 72.5

now we find the IQR which is Q3 - Q1.....= 72.5 - 20.5 = 52
now we multiply 52 by 1.5....52 * 1.5 = 78

so the outliers will be :
Q1 - 78 .....20.5 - 78 = -57.5...anything below this
Q3 + 78.....72.5 + 78 = 150.5...anything above this

there is no outliers in here

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A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich

Bess [88]

Answer:

a). k = 2.0794

b). $A_{t}=A_{0}(2^{3t} )$

c). 13107200 cells

d). rate of growth after 6 hours = 27255112

e). 4.76 hours

Step-by-step explanation:

From the formula of bacterial population,

$A_{t}=A_{0}e^{kt}$

Where $A_{t}$ = Bacterial population after time t

$A_{0}$ = Initial population

k = relative growth factor

t = duration or time

a). For $A_{t}=2\times 50=100$ cells

and $A_{0}=50$ cells

Time t = 20 minutes = $\frac{1}{3}$ hours

Now we plug in these values in the formula,

100 = $50(e^{\frac{k}{3}})$

$e^{\frac{k}{3}}=2$

By taking natural log on both the sides of the equation,

$\frac{k}{3}(lne)=ln(2)$

k = 3ln(2) = 2.0794

b). To get the expression we will plug in the value of k in the formula.

$A_{t}=A_{0}e^{kt}$

Since k = 3ln(2)

$A_{t}=A_{0}e^{3t(ln2)}$

Let y = $e^{3t(ln2)}$

By taking natural log on both the sides of the equation,

lny = $ln(e^{3t(ln2)} )$

lny = 3t(ln2)ln(e)

ln(y) = 3t(ln2)

ln(y) = $ln(2)^{3t}$

y = $2^{3t}$

Now our expression will be $A_{t}=A_{0}(2^{3t} )$

c). Number of cells after t = 6 hours

$A_{6}=50(2^{3\times 6} )$

$A_{6}=50(2^{18})$

$A_{6}=50\times (262144)=13107200$

d). We can get the rate of growth by finding derivative of the expression

$\frac{d}{dt}(A_{t})=\frac{d}{dt}[A_{0}(e^{kt}})]$

= $A_{0}[ke^{kt}]$

Now $\frac{d}{dt}(A_{6})=k.A_{6}$

= 2.0794×13107200

= 27255112

e) From the given formula,

$A_{t}=A_{0}(2^{3t} )$

$1000000=50(2^{3t} )$

$2^{3t}=20000$

3t(ln2) = ln(20000)

t = $\frac{ln(20000)}{3(ln2)}$

t = $\frac{9.90349}{2.07944}=4.76$ hours

3 0

3 years ago

A tree is growing on a hill. In an attempt to make the tree grow straight up, a 12-foot-long wire is attached to a tree at a poi

melisa1 [442]

For this case what we must do is use the law of cosines.
We have then:
12 ^ 2 = 6 ^ 2 + 10 ^ 2 - 2 * (6) * (10) cos (x)
Clearing we have:
cos (x) = (12 ^ 2- (6 ^ 2 + 10 ^ 2)) / (- 2 * (6) * (10))
x = acos ((12 ^ 2- (6 ^ 2 + 10 ^ 2)) / (- 2 * (6) * (10)))
x = 93.82 degrees
Answer:
The tree is growing in relation to the hill at:
x = 93.82 degrees

3 0

4 years ago

Read 2 more answers

HELP ME PLEASE ASAP <br> What is the area of this polygon?

zepelin [54]

We can split this up into 1 trapezium and 2 triangles

Area = (3/2) * (6 + 7) + 1/2 * 3 * 6 + 1/2 * 3 * 4

= 19.5 + 9 + 6

= 34.5 answer

Its B

8 0

3 years ago

The length of a rectangle is seven inches more than its width. Its area is 78 square inches. Find the width and length of the re

Alekssandra [29.7K]

Answer:

Length: 23.38

Width: 3.34

Step-by-step explanation:

First I drew out a diagram so that I could visibly see the problem. [ a rectangle with 7x on one side and x on the other.

So to find the area you multiply length time width and 7x times x is 7x^2. The actual area is given: 78, so you set that equal to 7x^2.

Then you want to divide by 7 to get 11.14. Then you want to take the square root of 11.14 to get 3.338 which rounds to 3.34 {try not to round until the very end]. This gives you the width of the rectangle, so to find the length you would have to multiply x by 7 to get 23.38. When you multiply the length and width (3.34 x 23.38) you get 78.08 which can round to 78.

7x^2 = 78

x^2 = 78/7

x^2= 11.142...

x = sqrt(1.142...)

x = 3.338 --> 3.34 --> width

3.34 x 7 --> 23.38 --> length

3.34 x 23.38 --> 70.8

8 0

3 years ago

Read 2 more answers

A livestock pen is built in the shape of a rectangle that is 2 times as long as it is wide. The perimeter is 30 feet. If the mat

shutvik [7]