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Mazyrski [523]
3 years ago
5

What is the ratio 15 :24 written in the lowest terms

Mathematics
2 answers:
Charra [1.4K]3 years ago
5 0
15:24 to the lowest term is 5:8. This is because you divide each number by the same number. You will divided 15 and 24 by 3 to get 5:8.
Sonbull [250]3 years ago
3 0
15:24

Divide 3 each:  15 divided by 3=5

24 divided by 3= 8

You can't go further than that.

Therefore your answer will have to be 5:8 or 5/8. Hope this helped!

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100% probability. the probability of an event is 1 (or 100%) if and only if the event can never occur.
Ierofanga [76]
True us the answer too your question.
8 0
4 years ago
The solution to 2x2 – 11 = 87 is
Ilia_Sergeevich [38]
2x^2 - 11 = 87
2x^2 = 87 + 11 = 98
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7 0
3 years ago
Read 2 more answers
The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth
Art [367]

Answer:

a) N_0=250\; k=0.15

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies t=0. And N(0)=N_0. The process is:

N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250

b) After 2 days means t=48. So, we just replace and operate:

N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}

c) N(t_1)=4000; \;t_1=?

N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\  \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h

d) t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)

N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h

6 0
3 years ago
YOOO MARKIN BRAINIEST !!
Brilliant_brown [7]
Y ≥ 2x-5
y-intercept is -5, from there go 2 up 1 over ( slope is rise over run)
greater than or equal so the line is connected
now find the shaded area by plugging in (0,0)
0 ≥2(0)-5
0 ≥-5 is correct so shade the area that to the left of the line, (the whole area that including (0,0))

y<-3x
y intercept is 0 so start from there and go down 3 right 1 (or go up 3 left 1)
broken like cause no or equal sign
the (0,0) is on the line so use (1,1) to find the answer
1<-3(1) is incorrect so shade the area that dies not include (1,1) or the entire area to the left of that line

you can see the section where both shaded area cross, thats the answer so erase every area you shaded that isn’t the answer so
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3 0
4 years ago
Read 2 more answers
Someone help with this
elena-14-01-66 [18.8K]

The quadrants are I, II, III, and IV, (meaning 1, 2, 3, and 4). The first quadrant is in the upper right, which has both positive x and positive y values. The second quadrant is the upper left, which has negative x and positive y values. The third quadrant is the lower left, which has both negative x and y values. The fourth quadrant is the lower right, which has positive x and negative y values. Using this knowledge and our positive and negative signs, the following are the answers to questions 1-6.

1) (-4, -2) - Quadrant III, both x and y are negative

2) (0, -7) - This point is actually on the y-axis. The x value is 0 and the y value is -7, so the graph is 7 units down from the origin on the y-axis.

3) (0,0) - This point is the origin, or where the x and y axes cross (the middle of the graph).

4) (6, -9) - This point is in Quadrant IV, because it has a positive x value and a negative y value

5) (3,5) - This point is in Quadrant I, because both the x and y values are positive.

6) (8,0) - This point is on the x-axis. The y-value is zero, so this point is 8 units to the right of the origin between Quadrant I and Quadrant IV.

Using the knowledge presented above, to graph the points given to you in the second part of the problem, first you can figure out what quadrant or part of the graph the point is on. Then, you can count the number of units (squares on the graph) in the right direction (remember that up is positive on the y-axis and down is negative, and to the right is positive on the x-axis and to the left is negative) in order to plot the points. Then, you must connect the points that correspond to the same figure in order to create the figures.

Please comment if you have any questions!

Hope this helps!

7 0
4 years ago
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