What is the the translation rule

In a right triangle, <A and <B are acute. Find the value of sin A, when tan A = 8/15

finlep [7]

Answer:sinA=8/17

Step-by-step explanation:

Using Pythagoras principle

x=√(15^2+8^2)

x=√(15x15+8x8)

x=√(225+64)

x=√(289)

x=17

SinA=opposite/hypothenus

SinA=8/17

3 0

2 years ago

What is the converse of the statement?<br> "If x - 2 = 5, then x = 7"

dlinn [17]

ANSWER

"If x=7, then x - 2 = 5"

EXPLANATION

Let

$p \to \: q$

be a propositional statement.

The converse of this statement is

$q \to \: p$

In other words, the converse of the statement,

"If p then q" is "If q, then p"

The given given conditional statement is

"If x - 2 = 5, then x = 7"

Therefore the converse is

"If x=7, then x - 2 = 5"

3 0

3 years ago

What are the coefficient(s) in the expression 6y+3x-2

Maslowich

Answer:

6 and 3

Step-by-step explanation:

3 0

2 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago

A research scientist wants to know how many times per hour a certain strand of bacteria reproduces. The mean is found to be 9.8

svetlana [45]

Answer:

The 99% confidence interval for the true mean number of reproductions per hour for the bacteria is between 9.6 and 10.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575\frac{2.4}{\sqrt{955}} = 0.2$

The lower end of the interval is the sample mean subtracted by M. So it is 9.8 - 0.2 = 9.6 reproductions per hour.

The upper end of the interval is the sample mean added to M. So it is 9.8 + 0.2 = 10 reproductions per hour.

The 99% confidence interval for the true mean number of reproductions per hour for the bacteria is between 9.6 and 10.

8 0

2 years ago