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3 years ago

Pleaseeee help meeeeeee

Mathematics

1 answer:

Allisa [31]3 years ago

8 0

Take your choice of the trig function you would like to use.
∠B = arcsin(24/26) = arccos(10/26) = arctan(24/10) ≈ 67.38°

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Sorry bout that, THIS IS THE NEW ONE PLEASE ANSWER

Sergio039 [100]

Answer:

Step-by-step explanation:

Putting value of y = 5

4(5)2 - 7(5) - 6

4(25) - 35 - 6

100 - 41

3 0

3 years ago

What is f (x) = 4x + 1; {–4, –2, 0, 2, 4}

Reika [66]

Answer

4x+1

−4,−2,0,2,4

Step-by-step explanation:

5 0

2 years ago

What is -3/8, 5/16, -0.65, 2/4 from least to greatest

storchak [24]

-3/8, -0.65, 5/16, 2/4

8 0

3 years ago

NEED HELP FAST PLEASE Evaluate the limit. a.4 c.1 b.2 d.limit does not exist

BlackZzzverrR [31]

We have the following limit:
(8n2 + 5n + 2) / (3 + 2n)
Evaluating for n = inf we have:
(8 (inf) 2 + 5 (inf) + 2) / (3 + 2 (inf))
(inf) / (inf)
We observe that we have an indetermination, which we must resolve.
Applying L'hopital we have:
(8n2 + 5n + 2) '/ (3 + 2n)'
(16n + 5) / (2)
Evaluating again for n = inf:
(16 (inf) + 5) / (2) = inf
Therefore, the limit tends to infinity.
Answer:
d.limit does not exist

7 0

3 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

3 years ago