6 cm

A PHN is studying the rate of H1N1 in the community. If there are five cases of H1N1 in the community of 1,000 people, what is t

BigorU [14]

Answer:

0.5%

Step-by-step explanation:

The prevalence rate shows the proportion of people that have a disease in a population and to be able to calculate it, you have to divide the number of people that have a disease by the population:

Prevalence rate=(5/1,000)*100

Prevalence rate=0.005*100

Prevalence rate=0.5%

According to this, the answer is that if there are five cases of H1N1 in the community of 1,000 people, the prevalence rate is 0.5%.

8 0

3 years ago

On 31st December 2015 the GBP/USD exchange rate was 1.47360. On 31st December 2016 it was 1.23321. Calculate the

Mrrafil [7]

16%
New/old-1
The pound depreciated 16 percent relative to the dollar

8 0

2 years ago

Square root of 363 - 3 times the square root of 27

SOVA2 [1]

129,582 or press windows key on the keyboard and type cal and press enter

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3 0

2 years ago

Find the distance between lines 8x−15y+5=0 and 16x−30y−12=0.

AlladinOne [14]

Answer:

The distance between the two parallel lines is 11/17 units

Step-by-step explanation:

we have

8x-15y+5=0 -----> equation A

16x-30y-12=0 ----> equation B

Divide by 2 both sides equation B

8x-15y-6=0 ----> equation C

Compare equation A and equation C

Line A and Line C are parallel lines with different y-intercept

step 1

Find the slope of the parallel lines (The slope of two parallel lines is the same)

8x-15y+5=0

15y=8x+5

$y=\frac{8}{15}x+\frac{1}{3}$

the slope is

$m=\frac{8}{15}$

step 2

Find the slope of a line perpendicular to the given lines

Remember that

If two lines are perpendicular then their slopes are opposite reciprocal (the product of their slopes is -1)

$m1*m2=-1$

we have

$m1=\frac{8}{15}$

therefore

$m2=-\frac{15}{8}$

step 3

Find the equation of the line perpendicular to the given lines

assume any point that lie on line A

$y=\frac{8}{15}x+\frac{1}{3}$

For $x=0$

$y=\frac{1}{3}$

To find the equation of the line we have

$point\ (0,1/3)$ ---> is the y-intercept

$m=-\frac{15}{8}$

The equation in slope intercept form is

$y=-\frac{15}{8}x+\frac{1}{3}$ -----> equation D

step 4

Find the intersection point of the perpendicular line with the Line C

we have the system of equations

$y=-\frac{15}{8}x+\frac{1}{3}$ ----> equation D

$8x-15y-6=0$ ----> $y=\frac{8}{15}x-\frac{2}{5}$ ----> equation E

equate equation D and equation E and solve for x

$\frac{8}{15}x-\frac{2}{5}=-\frac{15}{8}x+\frac{1}{3}$

$\frac{8}{15}x+\frac{15}{8}x=\frac{1}{3}+\frac{2}{5}$

Multiply by 120 both sides to remove fractions

$64x+225x=40+48$

$289x=88$

$x=88/289$

Find the value of y

$y=-\frac{15}{8}(88/289)+\frac{1}{3}$

$y=-\frac{206}{867}$

the intersection point is $(\frac{88}{289},-\frac{206}{867})$

step 5

Find the distance between the points $(0,\frac{1}{3})$ and $(\frac{88}{289},-\frac{206}{867})$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$d=\sqrt{(-\frac{206}{867}-\frac{1}{3})^{2}+(\frac{88}{289}-0)^{2}}$

$d=\sqrt{(-\frac{495}{867})^{2}+(\frac{88}{289})^{2}}$

$d=\sqrt{(\frac{245,025}{751,689})+(\frac{7,744}{83,521})}$

$d=\sqrt{\frac{314,721}{751,689}}$

$d=\frac{561}{867}\ units$

Simplify

$d=\frac{11}{17}\ units$

therefore

The distance between the two parallel lines is 11/17 units

see the attached figure to better understand the problem

4 0

3 years ago

If BE−→ B E → bisects ∠ABD and m∠ABE = 62°, find m∠ABD

Bas_tet [7]

Given:
Line segment BE bisects ∠ABD as shown in the figure.

Therefore
m∠ABE = m∠DBE
and
m∠ABD = m∠ABE + m∠DBE

Because m∠ABE = 62°, therefore m∠DBE = 62°, so that
m∠ABD = 62° + 62° = 124°

Answer: m∠ABD = 124°

4 0

3 years ago