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kkurt [141]
3 years ago
11

The black figure is an ellipse, and the black line segment is is major axis. What is the length of the blue line segment?

Mathematics
1 answer:
NeTakaya3 years ago
6 0

Answer:

D

Step-by-step explanation:

The formula we need to solve this is:

d_1+d_2=2a

Where

d_1  is the distance from one focus to the vertex

d_2  is the second distance from another focus to same vertex

<u>Note:</u> Here, vertex point is the point on top. The common point from which the RED and BLUE line meets the 2 foci

a  is half the length of the major axis (which is 9 here since total major axis is given as 18)

<em>Now we can plug into the equation and solve:</em>

<em>d_1+d_2=2a\\8.5+d_2=2(9)\\8.5+d_2=18\\d_2=18-8.5\\d_2=9.5</em>

<em />

<em>D is the correct answer.</em>

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melisa1 [442]
I used math

anyway
remember you can do anything to an equation as long as you do it to both sides

so

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2n+2 and 1/2=0+5
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minus 2 and 1/2 both sides
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divide both sides by 2
I mean times both sides by 1/2
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6 0
3 years ago
I'm a election 4713 people voted candidate B received 2 thirds of the votes. How many votes did candidate b receive
madam [21]

Answer:

3142

Step-by-step explanation:

4713 x 2/3

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3 years ago
Find the slope of a line that has points (-2,6) and (4,-1)
Nadya [2.5K]
-7/6 is the answer hope it helped
8 0
3 years ago
(-2) times (+5) times (-1) times (-5) times (-2)<br> please can u answer fast i have a test in 2 min
zlopas [31]

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Step-by-step explanation:

4 0
1 year ago
The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

6 0
3 years ago
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