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NARA [144]
3 years ago
13

Two places are 4.7 cm apart on a map.

Mathematics
1 answer:
Cloud [144]3 years ago
3 0
On the map , 
1cm=5km
therefore , 4.7cm= 5/1 x 4.7
=23.5 km <--------------answer
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Please help me with this :))
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Hey love! <3

Answer:

Your answer is 306 units^2!

Step-by-step explanation:

<em>Let's work this out together!</em>

<em>First let's find the top and bottom square's area.</em>

<em>6 x 15 = 90</em>

<em>Since there are not only two of those shapes but THREE, 90 x 3= 270.</em>

<em>So 270 is the surface area of the middle rectangles.</em>

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<em>Step 2! Let's find the area of these triangles. We know that both of the triangles are 6 on each side.</em>

<em>Think back to geometry formulas. The formula for finding the area of a triangle is different then a square.</em>

<em>Triangle formula: Base x Height / 2 = A</em>

<em>Rectangle formula: Base x Height = A</em>

<em>So 6 x 6 is 36, now we have to divide it by two; which equals 18.</em>

<em>Since there are TWO triangles, we multiply 18 by two giving us the total surface area of the triangles, 36.</em>

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<em>Now let's add our products together!</em>

<em>36 + 270 = 306</em>

<em>Thus the total surface area being </em><em>306 units^2</em>

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3 years ago
Two cards are selected at random without replacement from a well-shuffled deck of 52 playing cards. Find the probability of the
Vadim26 [7]
Consider the deck either as 4 suits of 13, or as 13 suits of 4. 

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3 0
4 years ago
Read 2 more answers
In a class 27 of students, 12 have a cat and 11 have a dog. There are 5 students who have a cat and a dog. What is the probabili
frosja888 [35]

Answer:

41.67% probability that a student has a dog given that they have a cat

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: having a cat.

Event B: having a dog.

12 of 27 students have a cat:

This means that P(A) = \frac{12}{27}

5 students who have a cat and a dog.

This means that P(A \cap B) = \frac{5}{27}

What is the probability that a student has a dog given that they have a cat?

P(B|A) = \frac{\frac{5}{27}}{\frac{12}{27}} = \frac{5}{12} = 0.4167

41.67% probability that a student has a dog given that they have a cat

6 0
4 years ago
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