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3 years ago

* During a 10 year period, the number of people living in Sherbury increased by 5% to 20 265

1 answer:

8 0

Answer:During 10years ago the number of people living in Sherbury increased by 965 while the number of people living in Yaston increased by 942

Step-by-step explanation:

During the 10 year period,

Let x= increase in the number of

people living in Sherbury

y=number of people living in Sherbury before the increase

Therefore,

5/100 * y = x

0.05y=x

And

x + y = 20265

But y=20x

x + 20x =20265

21x=20265

x=965

During 10years ago the number of people living in Sherbury increased by 965.

Likewise

Let

a=increase in the number of

people living in Yaston

b=number of people living in Yaston before the increase

7.5/100 * b = a

0.075b = a

b=13.33a

a + b = 13502

a + 13.33a=13502

14.33a=13502

a=942

During 10years ago the number of people living in Yaston increased by 942.

Conclusion

During 10years ago the number of people living in Sherbury increased by 965 while the number of people living in Yaston increased by 942.

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Answer:

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Answer:

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(-6x+3)+(5x-4)

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3 years ago

The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches. Men the same

masha68 [24]

Answer: The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.

Step-by-step explanation:

Let x and y area the random variable that represents the heights of women and men.

Given : The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches.

i.e. $\mu_1 = 64$ $\sigma_1=2.7$

Since , $z=\dfrac{x-\mu}{\sigma}$

Then, z-score corresponds to a woman 6 feet tall (i.e. x=72 inches).

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$z=\dfrac{72-64}{2.7}=2.96296296\approx2.96$

Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.

i.e. $\mu_2 = 69.3$ $\sigma_2=2.8$

Then, z-score corresponds to a man 5'10" tall (i.e. y =70 inches).

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3 years ago

Please explain the misleading

Ede4ka [16]

There are more compact cars (4*10 = 40) compared to trucks (2*10 = 20); however, the pictogram might make it appear that there are more trucks because the individual truck icon is larger compared to an individual compact car icon.

To anyone giving this image a quick glance, they may erroneously conclude that there are more trucks since their eye would notice the trucks first. Also, the person might think there are more trucks because bigger sizes tend to correspond to more proportion.

In real life, a truck is larger than a compact car, but the icons need to be the same size to have the figure not be misleading.

A very similar issue happens with the mid-size cars vs the compact cars as well. The three mid-size car icons span the same total width as the compact cars do, indicating that a reader might mistakenly conclude that there are the same number of mid-size cars compared to compact ones (when that's not true either).

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3 years ago

How many times longer is the wavelength of a sound wave with a frequency of 20 waves per second than the wavelength of a sound w

sergejj [24]

The sound wave with a frequency of 20 waves/sec is 800 longer than the wavelength of a sound wave with a frequency of 16,000 waves/sec

<h3>Calculating wavelength </h3>

From the question, we are to determine how many times longer is the first sound wave compared to the second sound water

Using the formula,

v = fλ

∴ λ = v/f

Where v is the velocity

f is the frequency

and λ is the wavelength

For the first wave

f = 20 waves/sec

Then,

λ₁ = v/20

For the second wave

f = 16,000 waves/sec

λ₂ = v/16000

Then,

The factor by which the first sound wave is longer than the second sound wave is

λ₁/ λ₂ = (v/20) ÷( v/16000)

= (v/20) × 16000/v)

= 16000/20

= 800

Hence, the sound wave with a frequency of 20 waves/sec is 800 longer than the wavelength of a sound wave with a frequency of 16,000 waves/sec

Learn more on Calculating wavelength here: brainly.com/question/16396485

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