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borishaifa [10]
3 years ago
12

A playground has a shape of a rectangle, with two semi-circles on it’s smaller sides as diameters added to its outside. If the s

ides of the rectangle are 38m and 28m, find the perimeter and the area of the playground
Using the pi formulae
Mathematics
1 answer:
Liula [17]3 years ago
4 0

Answer:

Area= 1371.72m^2

Perimeter: 251.84m

Step-by-step explanation:

A1=L*W

A1=38*28

A1=1064

A2=3.14*r^2

A2=3.14*196

A2=615.44/2

A2=307.72

A1+A2=1371.72

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How can you multiply 6 times 298 using distrubutive property
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1788 I did it in distributive property trust me
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Two objects, A and B, are connected by hinges to a rigid rod that has a length L. The objects slide along perpendicular guide ra
crimeas [40]
According to the questions statements it shows

y= L sin (theta)

and as it shows A and B are right triangle

y= sqrt (L^2 - x^2)

we need to find the speed of B by differentiating w.r.t time.
so,

d/dt = dx/dt . d/dx = -v d/dx

-v represents the A's velocity
so,
Vb = d/dt (y) = -v d/dx sqrt(L^2-x^2)
= v.x/ sqrt(L^2-x^2)

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3 0
3 years ago
A piñata is a container filled with toys and candy and is broken open by hitting it with a stick. Sophia is trying to break the
belka [17]

Answer:

A) 0.24

B) The probability mass function of X, the number of hits required to break the piñata

X | P(X)

0 | 0.00

1 | 0.70

2 | 0.24

3 | 0.054

4 | 0.006

Step-by-step explanation:

Probability of Sophia breaking the piñata on the first attempt = 0.7

Probability of Sophia NOT breaking the piñata on the first attempt = 1 - 0.7 = 0.3

Probability of Sophia breaking the piñata on the second atrempt = 0.8

Probability of Sophia NOT breaking the piñata on the second atrempt = 1 - 0.8 = 0.2

Probability of Sophia breaking the piñata on the third atrempt = 0.9

Probability of Sophia NOT breaking the piñata on the second atrempt = 1 - 0.9 = 0.1

Probability of Sophia breaking the piñata on the fourth attempt = 1.0 (this is the highest number of attempts as a probability 1.0 means that the piñata breaks on the fourth attempt if it hasn't broken by now)

A) The probability that Sophia does not break the piñata on the first hit and does break the piñata on the second hit.

The required probability = (Probability that Sophia does not break the piñata on the first hit) × (Probability that Sophia does break the piñata on the second hit)

= 0.3 × 0.8 = 0.24

B) Let the random variable X represent the number of hits required for Sophia to break the piñata. Complete the probability distribution of X in the table below Probability of x 0.7

- X = 0, P(X) = 0

- X = 1

Probability of Sophia breaking the piñata on the first hit = 0.7

- X = 2

Probability of Sophia not breaking the piñata on the first hit, but breaking it on the second hit = 0.3 × 0.8 = 0.24

- X = 3

Probability of Sophia not breaking the piñata on the first and second hit, but breaking it on the third hit = 0.3 × 0.2 × 0.9 = 0.054

- X = 4

Probability of Sophia not breaking the piñata on the first, second and third hit, but breaking it on the fourth hit = 0.3 × 0.2 × 0.1 × 1.00 = 0.006

The probability mass function is then

X | P(X)

0 | 0.00

1 | 0.70

2 | 0.24

3 | 0.054

4 | 0.006

To check of we are correct, the probabilities should sum up to give 1.0

The cumulative probability

= 0.00 + 0.70 + 0.24 + 0.054 + 0.006 = 1.00

Hope this Helps!!!!

8 0
3 years ago
(GEOMETRY E2020) What is the value of X?
Stella [2.4K]
Use the formula h^{2} =xy so:
x^{2} =9(16)
x^{2} =144
\sqrt{ x^{2} } = \sqrt{144}
x=12

4 0
3 years ago
Read 2 more answers
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