In ΔEFG, e = 690 inches, ∠F=94° and ∠G=46°. Find the length of g, to the nearest inch.

Decrease 950 by 36%.

horsena [70]

Answer:

342

Step-by-step explanation:

950×0.36=342.....

6 0

3 years ago

Read 2 more answers

The circle has a circumference of 9.42 units. What is the area of the square ? Use 3.14 for pi. Explain how you got your answer

Elan Coil [88]

So 9.42 is the circumference 9.42/3.14 is 3 and 3 is your diameter 1.5 is your radius and area is radius times radius then times pi so 1.5 x 1.5 = 2.25
2.25 x 3.14= 7.06500 so I believe your answer is about 7 units

3 0

2 years ago

What are the possible answers for the equation 9x?- 16 = 0 ? A. 1-13, 13) 4 4 } 3 3 C. 14, VA) D. 3, 3) 16 E.

Genrish500 [490]

Answer:

13

Step-by-step explanation:

because I did the math and I learned this today it's simple

3 0

2 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

Can someone explain this please???

xeze [42]

A function is a rule that assigns exactly one output to a given input. The input is taken from a set called the domain, and the corresponding output belongs to a set called the range.

1. In this exercise, we're calling the pool of patients 1-8 the domain, and the pool of nurses A-D the range. The given table describes a function because any patient is assigned to only one nurse.

2. This wouldn't be a function if at least one patient was assigned to more than one nurse. If this were to happen in practice, the patient could be, say, given the same dose of some medicine twice if the nurses aren't careful.

3. Making the nurse pool the domain and the patient pool the range would give a relation that is not a function, since more than one patient is assigned to one nurse.

4 0

3 years ago