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2 years ago

Find the circumference of the circle. Use 3.14 for pi. Just want to make sure my answer for question 7 is right.

Mathematics

1 answer:

leva [86]2 years ago

4 0

The circumference is 2PieR, so we'd multiply 1.65 by 2, and get 3.3.
Afterwards, we multiply 3.3 by 3.14, which is 10.362. Your answer is incorrect.

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Since 1990, the number of gallons of whole milk each person in the United States drinks has decreased 4.5% each year. In 1990, e

MariettaO [177]

Answer:

11.2 gallons

Step-by-step explanation:

by multiplying 16 by 0.965 (as it is a 4.5% decrease per year) and then multiplying the answer by 0.965 and so on until done a total of 10 times gets you the answer 11.2 gallons. there is a quicker way to do it but I cannot remember sorry

3 0

3 years ago

Read 2 more answers

Can somebody please help me with question number 1

inysia [295]

Answer:

1a: 6.283

b: 51.313

c: 285.1

Step-by-step explanation:

Area of a sector = (angle/360)πr²

Area=(45/360)π(4²)=6.283

Area=(120/360)π(7²)=51.313

Area=(270/360)π(11²)=285.1

6 0

2 years ago

How do i simplify -9x^3+8x^2-(7x/6)-(-7)

cricket20 [7]

Multiply -1 by -7
-9x^3 + 8x^2 - 7x/6 + 7

7 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: