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3 years ago

6

The distance between Vancouver and Winnipeg is approximately 1850 km in a straight line. The distance on a map is 3.7 cm. Write

a scale statement for the map. What scale factor was used to make the map.

1 answer:

yKpoI14uk [10]3 years ago

8 0

3.7cm on the map represents 1850km in reality

3.7cm : 1850km

1cm : 1850/3.7 km

1cm : 500km

Scale statement for map is : 1cm : 500km.

That is 1cm on map represents 500km.

1cm : 500km. Recall 1km = 1000m = 100 000cm

1cm : 500* 100000cm

1cm : 50 000 000cm

1: 50 000 000.

Scale factor is 50 000 000.

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The number of mosquitoes active near a pond is given by the function f(x)=32x−2 , where x is the number of minutes after dusk

galina1969 [7]

Answer:

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Step-by-step explanation:

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3 years ago

A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a

enyata [817]

Complete Question

A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a sample of 18 adults.

a-1. How many of the 18 would we expect to indicate that the Treasury should continue making pennies

a-2) What is the standard deviation?

a-3) What is the likelihood that exactly 3 adults would indicate the Treasury should continue making pennies?

Answer:

a-1 $\= x = 11 .52$

a-2 $\sigma = 2.036$

a-3 $P(3) = 4.7*10^{-5}$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 18$

The proportion of adult that felt the U.S. Treasury should continue making pennies is p = 0.64

The proportion of adult that feel otherwise is

$q = 1- p = 1-0.64 = 0.36$

The mean is mathematically evaluated as

$\= x = n * p$

substituting values

$\= x = 18 * 0.64$

$\= x = 11 .52$

The standard deviation is mathematically represented as

$\sigma = \sqrt{ npq}$

substituting values

$\sigma = \sqrt{18 * 0.64 * 0.36}$

$\sigma = 2.036$

The likelihood that 3 adult would indicate the Treasury should continue making pennies is mathematically evaluated as

$P(3) = \left n} \atop \right. C_3 (p)^{3} * (q)^{n-3}$

Now

$\left n} \atop \right. C_3 = \frac{n! }{[n-3] ! 3!}$

substituting values

$\left n} \atop \right. C_3 = \frac{18! }{[15] ! 3!}$

$\left n} \atop \right. C_3 = \frac{18 * 17 * 16 * 15! }{[15] ! (3 *2 *1 )}$

$\left n} \atop \right. C_3 = \frac{18 * 17 * 16 }{ (3 *2 *1 )}$

$\left n} \atop \right. C_3 = 816$

So

$P(3) = 816 * (0.64 )^3 * (0.36 )^{18-3}$

$P(3) = 4.7*10^{-5}$

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3 years ago

If a person can see 6 miles from a height of 25 feet, how far can a person see from a height of 196 feet?

Aleksandr-060686 [28]

Answer:

47,04 miles

Step-by-step explanation:

you can clearly see something 20 feet away that you should be able to see from that distance. If a person can see 6 miles from a height of 25 feet, by dividing 196 by 25 we can get the following calculation

196 ÷ 25 = 7,84

Then you gan multiply

7,84 × 6 = 47,04 miles

Or equal to 75703,542 meters

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3 years ago

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$ .

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

6 0

3 years ago