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vodka [1.7K]
2 years ago
10

How to figure a composite figure area

Mathematics
1 answer:
photoshop1234 [79]2 years ago
8 0

Answer:

To find the area of a composite figure or other irregular-shaped figure, divide it into simple, nonoverlapping figures. Find the area of each simpler figure, and then add the areas together to find the total area of the composite figure.

Step-by-step explanation:

Did this help, do you need an example

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Write the equation for the following table:<br> X. Y. <br> 0 3<br> 1 7<br> 2 11<br> 3 15
Vanyuwa [196]

Answer:

y = 4x + 3

Step-by-step explanation:

The equation of the table of values given can be written as y = 4x + 3 .

When x = 0,

y = 4(0) + 3

y = 0 + 3 = 3 this corresponds with what we have in the table above.

Let's try another.

When x = 2,

y = 4(2) + 3

y = 8 + 3

y = 11

This also corresponds with what we have in the table.

Therefore, the equation of the table given is y = 4x + 3 .

4 0
2 years ago
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A rectangular shaped park contains two gardens of multi-colored roses. Sidewalks enclose the whole park and each of the gardens.
astraxan [27]

Answer:

148 meters

Step-by-step explanation:

4 0
2 years ago
How to solve the problem
Dima020 [189]
A=1 B=2 C=3 Yes it would work for all because its using the commutative property.
5 0
2 years ago
Assume that a policyholder is four times more likely to file exactly two claims as to file exactly three claims. Assume also tha
kykrilka [37]

Answer:

1.3125

Step-by-step explanation:

Given that our random variable X follows a Poisson distributionP(X=k)=\frac{\lambda^k e^-^p}{k!} \ \ \ \ \ \ \ \ p=\lambda

Evaluate the formula at k=2,3:

P(X=2)=0.5\lambda^2e^{-\lambda}\\\\P(X=3)=\frac{1}{6}\lambda^2e^{-\lambda}

#since

4P(X=2)=P(X=3);\\\\0.5\lambda^2e^{-\lambda}=4\times\frac{1}{6}\lambda^3e^{-\lambda}\\\\0.5\lambda^2=\frac{2}{3}\lambda^3\\\\0.75=\lambda

The mean and variance of the Poisson distributed random variable is equal to \lambda:

\mu=\lambda=0.75\\\sigma ^2=\lambda=0.75

#By property variance:

\sigma ^2=V(X)=E(X^2)-(E(X))^2=E(X^2)\\\\E(X^2)=\sigma^2+\mu^2=0.75+0.75^2=1.3125

The expectation is 1.3125

4 0
3 years ago
Plz help I’ll mark you
timurjin [86]

Answer:

A 1/2

Step-by-step explanation:

Ratio of short length to hypotenuse

= cos60

= 1/2

4 0
3 years ago
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