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3 years ago

5

The population of the town increased from 175000 to 262500 in a decade. What is the average percent increase of population per y

ear?

1 answer:

Alenkinab [10]3 years ago

3 0

Answer:

5%

Step-by-step explanation:

As given,

The population of the town increased from 175000 to 262500 in a decade.

⇒ In 10 years , population increased from 175000 to 262500

⇒% increase = $\frac{262500 - 175000}{175000}$ × 100% = $\frac{87500}{175000}$ ×100% = 0.5×100% = 50%

Now,

Average % increase = $\frac{percent increase}{Total years}$ = $\frac{50}{10}$ % = 5%

∴ we get

Average percent increase of population per year = 5%

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Find the distance traveled in 35 seconds by an object traveling at a velocity of v(t) = 20 + 5cos(t) feet per second

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3 years ago

1. Vanessa and Renita are having a friendly competition at work. Their job is to unpack boxes. Vanessa has unpacked 3 fewer than

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The number of boxes renita and Vanessa has unpacked each is 6 and 9 boxes respectively.

<h3>Equation</h3>

Number of boxes renita has unpacked = x
Number of boxes Vanessa has unpacked = 2x - 3
Total boxes both unpacked = 15

x + (2x - 3) = 15

x + 2x - 3 = 15

3x = 15 + 3

3x = 18

x = 18/3

x = 6

Therefore,

Number of boxes renita has unpacked = x

= 6 boxes

Number of boxes Vanessa has unpacked = 2x - 3

= 2(6) - 3

= 12 - 3

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1 year ago

What multiplies to get -2 and adds to get -5

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3 years ago

6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP

adelina 88 [10]

Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = $360^{o}$ - $324^{o}$

= 0 $36^{o}$

Sum of angles in a triangle = $180^{o}$

<P + <Q + <S = $180^{o}$

048° + 0 $36^{o}$ + <S = $180^{o}$

$84^{o}$ + <S = $180^{o}$

<S = $180^{o}$ - $84^{o}$

= $96^{o}$

<S = $96^{o}$

Applying the Sine rule,

$\frac{y}{Sin P}$ = $\frac{x}{Sin Q}$ = $\frac{z}{Sin S}$

$\frac{y}{Sin P}$ = $\frac{z}{Sin S}$

$\frac{y}{Sin 48^{o} }$ = $\frac{17}{Sin 96^{o} }$

$\frac{y}{0.74314}$ = $\frac{17}{0.99452}$

⇒ y = $\frac{12.63338}{0.99452}$

= 12.703

y = 12.70 km

$\frac{x}{Sin Q}$ = $\frac{z}{Sin S}$

$\frac{x}{Sin 36^{o} }$ = $\frac{17}{Sin 96^{o} }$

$\frac{x}{0.58779}$ = $\frac{17}{0.99452}$

⇒ x = $\frac{9.992430}{0.99452}$

= 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

6 0

3 years ago