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Ksivusya [100]
3 years ago
11

P l e a s e h e l p ! ! !

Mathematics
2 answers:
Fiesta28 [93]3 years ago
4 0

Answer:

3.35

Step-by-step explanation:

Can’t be shorter than the sides I don’t think and 11 is way to high

storchak [24]3 years ago
3 0
Correct answer to this would be 3.35
You might be interested in
Which shows the expressions in the order they would appear on a number line from least to greatest?
Ganezh [65]
2^3 = 8
11/9 = 1.22
sqrt 5 = 2.24
sqrt 20 = 4.47
sqrt 11 = 3.32

least to greatest : 11/9, sqrt 5 , sqrt 11 , sqrt 20, 2^3
7 0
3 years ago
Help me I want good grades i
Anna71 [15]
The second one the unit rate is 2.05 i think
5 0
3 years ago
Sylvia wants to save up $1375 to go on vacation to the beach and amusement parks. She saves $55 each week and now has $165 saved
laiz [17]

Answer:

Step-by-step explanation:

55x + 165 = 1375

so just add 55 till you get there

22

6 0
4 years ago
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:

\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)

And for the next step we have:

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

And with this we have the requiered proof.

And since b=1-p we have that:

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

4 0
4 years ago
GUYS PLZ HELP ME!!!!!!
siniylev [52]

Answer:

B E F G

Step-by-step explanation:

Polygons are 2 dimensional shapes with straight edges (no curved edges). The shape needs to be closed up entirely to be a polygon. It cannot have any intersecting line segments

A can't be a polygon because the shape isn't closed up entirely

C can't be a polygon because it is a 3d shape

D can't be a polygon because it has curved edges.

H can't be a polygon because it isn't closed up (there are no line segments).

I can't be a polygon because polygons cannot have intersecting lines

J can't be a polygon because polygons cannot have intersecting lines

I hope this helps! If it does, a rate or brainliest would be much appreciated. :)

7 0
3 years ago
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