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3 years ago

8

The base of a rectangular prism has a length of 13 inches and a width of 1over2. The volume of the prism is less than 65 cubic i

nches. Find all possible heights of the prism.

1 answer:

Vaselesa [24]3 years ago

8 0

The answer might be 26

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How do i do sohcahtoha

DochEvi [55]

Answer:

you mean trigonometry i'm assuming

but essentially:

soh = sin theta (the angle) = opposite length/hypotenuse length

cah = cos theta (angle) = adjacent length/hypotenuse length

toa = tan theta (angle) = opposite length/adjacent length

Step-by-step explanation:

8 0

2 years ago

Could someone help me understand this?

larisa86 [58]

Answer:

8 < x < 40

Step-by-step explanation:

x − 8 must be more than 0, but it can't be greater than 32.

0 < x − 8 < 32

8 < x < 40

A more precise answer would require law of cosines and calculus.

5 0

3 years ago

How to solve 2.2z=6.8

Fudgin [204]

Answer:

3.09

Step-by-step explanation:

divide 2.2 by 6.8 then z would equal 3.09

4 0

3 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:

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

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

3k to the 2 power - 10k - 7

makkiz [27]

Answer:

Use a formulae calculator on Google

7 0

2 years ago