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3 years ago

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

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