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navik [9.2K]
3 years ago
8

How do maps use the coordinate plane for locating towns?

Mathematics
1 answer:
Degger [83]3 years ago
7 0

Answer:

Your street name and house number furnish your absolute location. Latitude and longitude provide absolute location for a place on the globe. We can locate places on the globe by determining where lines on latitude and longitude cross.

Step-by-step explanation:

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John wrote the numbers 678,901 and 67,890. How many times greater is the value of 7 in 678,901?​
worty [1.4K]

Answer:

The value of 7 in 678,901 is 10 times greater than the value of 7 in 67,890

Step-by-step explanation:

step 1

we have

678,901

Write in expanded form

678,901=600,000+70,000+8,000+900+1

The value of 7 is 70,000

step 2

we have

67,890

Write in expanded form

67,890=60,000+7,000+800+90

The value of 7 is 7,000

step 3

Divide 70,000 by 7,000

\frac{70,000}{7,000}=10

therefore

The value of 7 in 678,901 is 10 times greater than the value of 7 in 67,890

4 0
3 years ago
Read 2 more answers
Josh wants to figure out how far he can run in 30 minutes or less. He averages 8 minutes to run a mile. What inequality can he u
OLga [1]
Time to run 1 mile * number of miles run must be 30 minutes or less
Time to run a mile is 8
Used m = number of miles ran
8m <= 30

<= is less than or equal to
8 0
3 years ago
Let X be a continuous random variable with density functionf(x)={1−|x|0 for −1
BlackZzzverrR [31]

Answer:

f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1

Step-by-step explanation:

For this case we want to find the density function for Y=X^2

And we have the following density function for the random variable X:

f(X) = 1- |X|,-1 \leq X \leq 1

So we can do this replacing Y=X^2

F_Y (Y \leq y) = P(X^2 \leq y)

If we apply square root on both sides we got:

P(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_{-\sqrt{y}}^0 1+t dt +\int_{0}^{\sqrt{y}} 1-t dt

And if we integrate we got this:

F_Y (y) = [t+ \frac{t^2}{2}] \Big|_{-\sqrt{y}}^0+ [t -\frac{t^2}{2}] \Big|_{0}^{\sqrt{y}}

And replacing we got:

F_Y (y) = [0 -(-\sqrt{y} +\frac{y}{2})] + [\sqrt{y} -\frac{y}{2}]

F_Y (y) = 2 |\sqrt{y}| -y

And if we want to find the density function we just need to derivate the pdf like this:

f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1

3 0
4 years ago
Solve this equation <br> h(t)= -16t+175t+500
KiRa [710]
Use the quadratic formula where a = -16, b = 175 and c = 500 to find t. See picture for quadratic formula.

3 0
3 years ago
Can you please explain
hichkok12 [17]
The hole is x+4=0 because (x+4) is canceling out so the hole x=-4
Hopes this loves.
7 0
3 years ago
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