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Alexxx [7]
3 years ago
13

05.02)A doghouse is to be built in the shape of a right trapezoid, as shown below. What is the area of the doghouse?

Mathematics
1 answer:
jeka57 [31]3 years ago
3 0

Answer:

66.5 square feet

Step-by-step explanation:

See the given diagram attached with this question.

The right trapezoid consists of two parts, one square area having side length 7 ft and a right triangle having base 5 ft and altitude 7 ft.

Therefore, the area of the doghouse which is in shape of the given right trapezoid will be = area of the square + area of the triangle

= ( 7 × 7) + ( 0.5 × base × altitude )

= ( 7 × 7) + ( 0.5 × 5 × 7 )

= 49 + 17.5

= 66.5 square feet. (Answer)

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Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8
Delvig [45]
We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34
Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86

Thus the 98% confidence interval will be (105.34, 141.86)


Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
3 0
3 years ago
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UkoKoshka [18]

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2 years ago
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Bill and Eugene left the airport at the same time. They traveled in opposite directions. Eugene traveled 21.1 km/h faster than B
gizmo_the_mogwai [7]

Answer:

56.3 kilometers per hour

Step-by-step explanation:

We know D = RT

where

D is distance

R is rate (speed)

T is time

Let's Bill's rate be "x". Thus using equation we can write:

D = xt

We know Eugene travelled 21.1 faster than Bill, so Eugene's rate is 21.1 + x. So we can write:

D = (21.1 + x)t

D = 21.1t + xt

Now, putting t = 1 in both since after 1 hour given, we have:

D = x(1)

D = x

and

D = 21.1(1) + x(1)

D = 21.1 + x

We know the sum of both the distance is 133.7, so we can write:

x + 21.1 + x = 133.7

Solving for x:

2x = 133.7 - 21.1

2x = 112.6

x = 112.6 / 2

x = 56.3 km/hr

This is Bill's rate: 56.3 kilometers per hour

8 0
3 years ago
Convert 5.45787878 to a fraction. Simplify
TiliK225 [7]

Answer:

It is 54033/9900

Step-by-step explanation:

7 0
3 years ago
Problem1 The behavior of a physical system can be described by the following first order differential equation: dy/dt=2y +t^2
daser333 [38]

Answer:

y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}.

Step-by-step explanation:

Using first order linear differential equation:

\dfrac{\mathrm{d} y}{\mathrm{d} t} = 2y + t^2

\frac{\mathrm{d} y}{\mathrm{d} t} - 2y =t^2

finding integrating factor:

I.F = e^{\int -2dt}

I.F =e^{-2t}

now,

y = \dfrac{1}{IF}(\int t^2dt+ c )

y = \dfrac{1}{e^{-2t}}(\int t^2dt+ c )

y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}

hence the solution is

y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}

8 0
3 years ago
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