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3 years ago

What ratio is equivalent to 6:4

Mathematics

2 answers:

Lelechka [254]3 years ago

7 0

Forgive me if I'm wrong, but I think it is 3:2

Yanka [14]3 years ago

7 0

18:12
3:2
12:8
(i think i did it right)

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Help please! Identify an equation in point slope form for the line perpendicular to Y =-4x + 8 that passes through (-2, 7)

777dan777 [17]

Answer:

D. $y-7=\frac{1}{4}(x+2)$

Step-by-step explanation:

The point-slope form of a line is given by:

$y-y_1=m(x-x_1)$

The line given to us has equation; $y=-4x+8$ .

The slope of this line is -4. The of the line perpendicular to this line is the negative reciprocal of the slope of the given line.

Our slope of interest is therefore; $m=\frac{1}{4}$

Since the point goes through (-2,7), we have $x_1=-2,y_1=7$ .

We plug in the slope and the point into the point-slope formula to obain;

$y-7=\frac{1}{4}(x--2)$

The required equation is:

$y-7=\frac{1}{4}(x+2)$

3 0

3 years ago

A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t

11Alexandr11 [23.1K]

Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

$dm/m=0.2dt$

b) Integrate

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) Antilogarithm

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

Use the measured rate of 4 grams per hour after 3 hours

$t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

Particular solution:

$m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

3 0

3 years ago

If you draw a card with a value of 4 or less from a standard deck of cards, I will pay you 165$ if not your pay me $45. (Aces ar

jekas [21]

Answer: The required expected value is $3.46.

Step-by-step explanation:

Since we have given that

Number of value of 4 or less are

2,3,4

So, there are 12 in numbers.

So, probability would be

$\dfrac{12}{52}$

Since Aces are considered as the highest card in the deck.

So, remaining probability would be

$\dfrac{40}{12}$

Amount paid or value of 4 or less = $165

Amount not paid for other case = $45

So, the expected value would be

$\dfrac{12}{52}\times 165-\dfrac{40}{52}\times 45\\\\=\dfrac{1980}{52}-\dfrac{1800}{52}\\\\=\dfrac{180}{52}\\\\=3.46$

Hence, the required expected value is $3.46.

7 0

3 years ago

Assume that the Poisson distribution applies and that the mean number of aircraft accidents is 9 per month. Find P(0), the prob

Ugo [173]

Answer: 0.0001

It is unlikely to have a month with no aircraft accidents .

Step-by-step explanation:

Given : Mean number of aircraft accidents = 9 per month

The Poisson distribution formula :-

$\dfrac{e^{-\lambda}\lambda^x}{x!}$ , where $\lambda$ is the mean of the distribution.

If X = the number of aircraft accidents per month, then the probability that in a month, there will be no aircraft accidents will be :-

$\dfrac{e^{-9}(9)^0}{0!}=0.000123409804087\approx0.0001$

Hence, the probability that in a month, there will be no aircraft accidents = 0.0001

Since this is less than 0.5 , therefor it is unlikely to have a month with no aircraft accidents .

5 0

3 years ago

What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4

snow_tiger [21]

What is the third term of the sequence defined by the recursive rule f(1)=3 f(n)=f(n-1)+4

Need f(2):

f(2)=f(2-1)+4
f(2)=f(1)+4
f(2)=(3)+4=7

FIND f(3):
f(3)=f(3-1)+4
f(3)=f(2)+4
f(3)=(7)+4
f(3)=11

5 0

3 years ago