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2 years ago

Can anyone help with this:

Mathematics

1 answer:

boyakko [2]2 years ago

3 0

Maybe 3 goals ###################

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The area of a rectangular park is same as area of a square park. If the length

Leona [35]

Answer:

Step-by-step explanation:

Step one:

given data

length of rectangle= 24.5m

let the width be x

width of square= 14m

let the length be x

Step two:

we know that the area of a rectangle is

Ar= Length * width

Ar=24.5*x-----------1

Also, the area of the square is

As= Length * width

for square L=W

As= 14*14-----------------2

As=196m^2

equating 1 and 2 we have

14*14=24.5x

196=24.5x

divide both sides by 24.5

x=196/24.5

x=8m

<u>The Breadth/width of the rectangular park is 8m</u>

5 0

2 years ago

Ellen draws two plane figures.one of the figures has 3 more angles than the other figure.what shapes could she have drawn?

kifflom [539]

A triangle and pentagon.
Or. Quadrilateral and octagon

7 0

3 years ago

Can a right triangle be formed using these squares

Ilya [14]

Yes. The two smaller squares have a sum of 169 which is the value of the larger square.

a^2 + b^2 = c^2

25 + 144 = 169

It can be done. Notice the figure below shows you how to arrange the squares to give the answer of a^2 + b^2 = c^2

7 0

3 years ago

How many blocks are needed for the 5th example?

JulsSmile [24]

Answer:

5th example: 28 blocks.

24th example: 123 blocks.

Step-by-step explanation:

Hope this helped!

7 0

2 years ago

According to the Stack Overflow Developers Survey of 20184, 25.8% of developers are students.The probability that a developer is

murzikaleks [220]

Answer:

0.0326 = 3.26% probability that she is a student.

Step-by-step explanation:

Conditional Probability

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: Woman developer

Event B: Student

Probability that the developer is a woman:

7.4% of 25.8%(students).

76.4% of 100 - 25.8 = 74.2%(not students). So

$P(A) = 0.074*0.258 + 0.764*0.742 = 0.58598$

Student and woman developer.

7.4% of 25.8%(students), so

$P(A \cap B) = 0.074*0.258 = 0.019092$

If we encounter a woman developer, what is the probability that she is a student

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.019092}{0.58598} = 0.0326$

0.0326 = 3.26% probability that she is a student.

5 0

2 years ago