The legend on a map states that1 cm is 50 miles. If you measure
4 centimeters on the map, how
many miles would the actual
distance be?
Actual distance = [ ? ] miles

Answers

Answer 1
Answer:

Answer:

200 miles

Step-by-step-explination

1 x 4 = 4

4 x 50 = 200

200 miles.

Answer 2
Answer:

Answer:

200

Step-by-step explanation:


Related Questions

The square root of the sum of two consecutive odd integers is equal to 6. Find the larger integer.
M+ m + nm; use m = 6, and n = 4
Heather sampled the ages of 50 randomly selected people that frequented a store and obtained a mean of 24. She constructed a 95% confidence interval which was (19.1, 28.9). What is the margin of error?A) 2.5 years B) 4.9 years C) 5 years D) 6.7 years
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Dixon and his little sister Ariadne stand next to each other on the playground on a sunny afternoon. Their mother measures their shadows. Dixon's shadow is 18 feet long and Ariadne's shadow is 15 feet long. If Dixon is 6 feet tall, how tall is Ariadne?

Answers

Ariadne is 5 feet tall because 18 is 6x3 which would mean you would have to divide 15 by three to get the answer which is 5.
She would be 5ft tall because if you set the problem up in a ratio and cross multiple and divide 5 is your answer

A set of data includes 101 data points ranging from 0 – 200. If the data is split into classes with a range of 20 (1 – 20, 21 – 40, etc), and there are no values that fall in the class 101 – 120, what can you say about the data?A.The mean of the data will not be between 101 and 120.
B. All the data lies below 100.
C. The relative frequency of the class 101 – 120 is 0.
D. The median could be in the range 101 – 120.

Answers

Answer:

The relative frequency of the class 101 – 120 is 0.

Step-by-step explanation:

i took the quiz

D, because there might be values higher than 120, because the data points range from 0 - 200.

B(x) = 7x + 3 and h(x) = 9x - 1.
Find (b+h)(x)

Answers

(b+h)(x)=16x+2

Since we all know the theory that shows that a(x)+b(x)=(a+b)(x), we can use that in this problem. Simply simplify the problem and divide it into 2... b(x) plus h(x). This is simple as we already know the values of these variables so just add both and the sum will be the answer to your problem.

Find the area bounded by the curve y=x2 and the straight line y=2+xA.4 1/6
B. 4 1/2
C. 5 1/6
D. 5 1/2

Answers

x^2=2+x\n x^2-x-2=0\n x^2+x-2x-2=0\n x(x+1)-2(x+1)=0\n (x-2)(x+1)=0\n x=2 \vee =-1\n\n \displaystyle A=\int \limits_(-1)^22+x-x^2\, dx\n A=\left[2x+(x^2)/(2)-(x^3)/(3)\right]_(-1)^2\n A=2\cdot2+(2^2)/(2)-(2^3)/(3)-\left(2\cdot(-1)+((-1)^2)/(2)-((-1)^3)/(3)\right)\n A=4+2-(8)/(3)-\left(-2+(1)/(2)+(1)/(3)\right)\n A=6-(8)/(3)+2-(1)/(2)-(1)/(3)\right)\n A=8-(9)/(3)-(1)/(2)\n A=8-3-(1)/(2)\n A=5-(1)/(2)\n A=4(1)/(2)

Which is the best estimate of
11 1/5 and 2 3/4

Answers

I think your question lacks a little bit of information. After a little research, I am able to find that you probably wanted to say '11 1/5 divided by 2 3/4'.

Therefore, I am solving the question assuming that you want to say:

''What is the best estimate of 11 1/5 divided by 2 3/4''.

Answer:

The best estimate of  11 1/5 divided by 2 3/4 will be: 4

Step-by-step explanation:

Given the information

11 1/5 divided by 2 3/4

solving the expression

11(1)/(5)/ 2(3)/(4)

\mathrm{Convert\:mixed\:numbers\:to\:improper\:fraction:}\:a(b)/(c)=(a\cdot \:c+b)/(c)

so

  • 11(1)/(5)=(11\cdot 5+1)/(5)=(56)/(5)
  • 2(3)/(4)=(2\cdot 4+3)/(4)=(11)/(4)

so the expression becomes

11(1)/(5)/ 2(3)/(4)\:=(56)/(5)/ \:(11)/(4)

                =(56)/(5)* (4)/(11)

                =(56* \:4)/(5* \:11)

               =(224)/(5* \:11)

               =(224)/(55)

                =4.07

                = 4       (Rounded to the nearest 1 orthe Ones Place)

Therefore, the best estimate of  11 1/5 divided by 2 3/4 will be: 4

There are 77 students in the student council. The ratio of girls to boys is 7:4. How many girls are in the student council?

Answers

77 students
G : B
7 : 4

11 units (7+4) = 77
1 unit = 7

7 units = 49

so there are 49 girls in the student council

hope this helped :)