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Keith_Richards [23]
1 year ago
11

What is the base of a triangle with an area of 207 square inches and a height of 18 inches

Mathematics
1 answer:
White raven [17]1 year ago
4 0
Answer: 23 in

Step-by-Step Explanation:

Height (h) = 18 in
Base (b) = ?
Area (A) = 207 sq. in

We know,
Area of a Triangle = 1/2 * b * h

Therefore,
1/2 * b * h = 207
1/2 * b * 18 = 207
b * 9 = 207
9b = 207
b = 207/9
=> b = 23

Base (b) = 23 in
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A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t
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Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

a)

\frac{dy}{dt}\ \alpha\  y(1-y)

\frac{dy}{dt}=ky(1-y)

where k is the constant of proportionality, dy/dt =  rate at which the rumor spreads

b)

\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}}  =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\

y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days

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Step-by-step explanation:

You need to find

(f.g)=2 or (f.g) (-2)=2

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A sample of 15 commuters in Chicago showed the average of the commuting times was 33.2 minutes. If s = 8.3 minutes, find the 95%
OleMash [197]

Answer:

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Step-by-step explanation:

<u>Step :- (i)</u>

Given sample size 'n' =15

sample of the mean x⁻ = 33.2

The standard deviation of the sample 'S' = 8.3

<u>95% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

<u>Step:-(ii)</u>

<u>The degrees of freedom γ=n-1 = 15-1=14</u>

The tabulated value t = 1.761 at 0.05 level of significance.

now substitute all possible values, we get

(33.2 - 1.761\frac{8.3}{\sqrt{15} } ,33.2+ 1.761\frac{8}{\sqrt{15} } )

After calculation , we get

(33.2-3.7739 , 33.2+3.7739

(29.4261 ,36.9739)

<u>Conclusion</u>:-

the 95% confidence interval of the true mean.

(29.4261 ,36.9739)

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