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2 years ago

15) a tree casts a shadow 21 ft while Enzo who is 5 ft tall casts a shadow 6 ft long. (with picture)

Mathematics

1 answer:

Darina [25.2K]2 years ago

7 0

Set up a ratio between the tree and Enzo.

5 ft / 6ft = x/21 ft.

Cross multiply:

(5 x 21) = (6 x x)

Simplify:

105 = 6x

Divide both sides by 6:

x = 105 / 6

x = 17.5 feet tall.

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3 years ago

Four times the sum of three consecutive integers is 156.Write and solve an equation that will find the smallest integer.

Burka [1]

Answer:

4(x+x+1+x+2)=156, x=12

Step-by-step explanation:

let the numbers be x, x+1 and x+2.

Now,

4(x+x+1+x+2)=156

4(3x+3)=156

divide both side by 4

3x+3=156/4=39

subtract 3 from both sides

3x=39-3

3x=36

divide both sides by 3

x=36/3=12

x=12

Hence, the smallest of the integers is 12.

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have: