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Nesterboy [21]
2 years ago
5

Bobby knows that the perimeter of the original rectangle is 120 meters. He also knows that the perimeter of the reduced rectangl

e is 30 meters and the reduced length is 9 meters. A small rectangle has a length of 9 meters. A larger rectangle is blank. Not drawn to scale What is the width of the original rectangle? 20 meters 24 meters 36 meters 48 meters
Mathematics
2 answers:
Reika [66]2 years ago
7 0

Answer:

b 24 meters

Step-by-step explanation:

ed

Whitepunk [10]2 years ago
6 0

Answer:

<u>24 meters</u> is the width of the original rectangle.

Step-by-step explanation:

Given:

Bobby knows that the perimeter of the original rectangle is 120 meters. He also knows that the perimeter of the reduced rectangle is 30 meters and the reduced rectangle has a length of 9 meters.

Now, to get the width of original rectangle.

The reduced rectangle's perimeter = 30 meters.

The reduced rectangle's length = 9 meters.

Now, we find the width of reduced rectangle by using formula:

Let the width of reduced rectangle be x.

Perimeter=2\times length+2\times width

30=2\times 9+2\times x

30=18+2x

<em>Subtracting both sides by 18 we get:</em>

<em />12=2x<em />

<em>Dividing both sides by 2 we get:</em>

6=x\\\\x=6\ meters.

The width of reduced rectangle = 6 meters.

Now, to get the width of original rectangle:

Let the width of original rectangle be w.

<em>As given, the perimeter of the original rectangle = 120 meters.</em>

<em>And, the perimeter of reduced rectangle is 30 meters and its width is 6 meters.</em>

<em>So, 30 is equivalent to 6.</em>

<em>Thus, 120 is equivalent to </em>w.<em />

Now, to get the width using cross multiplication method:

\frac{30}{6}=\frac{120}{w}

<em>By cross multiplying we get:</em>

<em />30w=720<em />

<em>Dividing both sides by 30 we get:</em>

<em />w=24\ meters.<em />

<em>The width of original rectangle = 24 meters.</em>

Therefore, 24 meters is the width of the original rectangle.

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2 years ago
The number of airline passengers in 1990 was 466 million. The number of passengers traveling by airplane each year has increased
taurus [48]

Answer:

2010.

Step-by-step explanation:

We have been given an exponential growth formula P(t)=466\cdot 1.035^t, which represents number of passengers traveling by airplane since 1990.

To find the year in which 900 million passengers will travel by airline, we will equate the given formula by 900 and solve for t as:

900=466\cdot 1.035^t

\frac{900}{466}=\frac{466\cdot 1.035^t}{466}1.9313304721030043=1.035^t

Take natural log of both sides:

\text{ln}(1.9313304721030043)=\text{ln}(1.035^t)

Using property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

\text{ln}(1.9313304721030043)=t\cdot \text{ln}(1.035)

0.658209129198=t\cdot 0.034401426717

\frac{0.658209129198}{0.034401426717}=\frac{t\cdot 0.034401426717}{0.034401426717}

19.1331927775=t\\\\t=19.1331927775

This means that in the 20th year since 1990, 900 million passengers would travel by airline.

1990+20=2010

Therefore, 900 million passengers would travel by airline in 2010.

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1 year ago
Vince is saving for a new mobile phone. The least expensive model Vince likes costs $225.90. Vince has saved $122.35. He used th
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Answer:

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Step-by-step explanation:

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Let A(t) be the amount of salt in the tank at time t. We're given that A(0)=40. The rate at which this amount changes is given by

A'(t)=\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{4\text{ oz}}{1\text{ gal}}-\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ oz}}{100+(5-5)t\text{ gal}}

A'(t)+\dfrac{A(t)}{20}=20

e^{t/20}A'(t)+e^{t/20}\dfrac{A(t)}{20}=20e^{t/20}

\bigg(e^{t/20}A(t)\bigg)'=20e^{t/20}

e^{t/20}A(t)=400e^{t/20}+C

A(t)=400+Ce^{-t/20}

Since A(0)=40, we get

40=400+C\implies C=-360

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After 20 minutes, the tank contains

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2 years ago
A very large gardening business grows rose bushes for sale to garden stores around the world. The most popular colors are red, p
cricket20 [7]

Complete options are;

a. The approximation requires np > 10 and n(1 - p) > 10.

b. The sample size here is too small to use the Normal approximation to the binomial.

c. The approximation requires np > 30.

d. The Normal approximation works better if the success probability p is close to p = 0.5.

Answer:

Option C is false

Step-by-step explanation:

Looking at the options,

In normal approximation to the binomial,

n is the sample size,

p is the given probability.

q = 1 - p

Now, one of the conditions for using normal approximation to the binomial is that; np and nq or n(1 - p) must be greater than 10.

This means that option A is true because we require np or n(1 - p) to be greater than 10.

From Central limit theorem, the sample size needs to be more than 30 for us to use normal approximation. Our sample is 10. Thus, option B is true.

The approximation doesn't require np > 30. Rather it's the sample size that needs to be more than 30. Thus, option C is false.

Generally, when the value of p in a binomial is close to 0.5, the normal approximations will work better than when the value of p is closer to either 0 or 1. The reason is that: for p = 0.5, the binomial distribution will be symmetrical. Thus, option D is correct.

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2 years ago
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