Pls answer the question

Three highways connect city A with city B. Two highways connect city B with city C.

QveST [7]

(a) The probability that there is no open route from A to B is (0.2)^3 = 0.008.
Therefore the probability that at least one route is open from A to B is given by: 1 - 0.008 = 0.992.
The probability that there is no open route from B to C is (0.2)^2 = 0.04.
Therefore the probability that at least one route is open from B to C is given by:
1 - 0.04 = 0.96.
The probability that at least one route is open from A to C is:
$0.992\times0.96=0.9523$

(b)
α The probability that at least one route is open from A to B would become 0.9984. The probability in (a) will become: $0.9984\times0.96=0.95846$

β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:
$0.992\times0.992=0.9841$

Gamma: The probability that a highway between A and C will not be blocked in rush hour is 0.8. We need to find the probability that there is at least one route open from A to C using either a route A to B to C, or the route A to C direct. This is found by using the formula:
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$0.9523+0.8-(0.9523\times0.8)=0.99$
Therefore building a highway direct from A to C gives the highest probability that there is at least one route open from A to C.

4 0

2 years ago

Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units. Which statements bes

valentinak56 [21]

Answer:

<h3>Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.</h3><h3>The value of x is 8.</h3>

Step-by-step explanation:

Given that Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units

From the given Q is the middle point, which cut the diagonal PA into 2 equal halves.

By the definition of rhombus, diagonals meet at right angles.

Implies that PQ = QA

x+2 = 3x - 14

x-3x=-14-2

-2x=-16

2x = 16

dividing by 2 on both sides, we will get,

$x =\frac{16}{2}$

<h3>∴ x=8</h3><h3>Since Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles we can equate x+2 = 3x-14 to find the value of x.</h3>

The line segment $\overrightarrow{PA}=\overrightarrow{PQ}+\overrightarrow{QA}$

$\overrightarrow{PA}=x+2+3x-14$

$=4x-12$

$=4(8)-12$ ( since x=8)

$=32-12$

$=20$

<h3>∴ $\overrightarrow{PA}=20$ units</h3>

3 0

2 years ago

How does the area of triangle RST compare to the area of triangle LMN? The area of △RST is 2 square units less than the area of

sashaice [31]

Inscribe triangle RST in the square with dimensions 4×4, as shown in the figure.

from the area of this square, 4*4=16, we remove the triangles with dimensions
3×4, 2×1 and 2×4, whose side lengths are shown in the figure, and we are left with the area of triangle RST.

so $Area(RTS)=16- \frac{1}{2}*3*4- \frac{1}{2}*2*1- \frac{1}{2}*2*4=16-6-1-4=5$ units squared

similarly,

$Area(LMN)=4*3- \frac{1}{2}*1*4- \frac{1}{2}*2*2- \frac{1}{2}*2*3=12-2-2-3=5$ units squared

Thus, the areas are equal.

8 0

2 years ago

Read 2 more answers

If y is the circumcenter is angle STU find each measure

snow_tiger [21]

Answer:

Measures are SV=9 units., SY=14 units, YW= $\sqrt75units$ , YW= $\sqrt27units$

Step-by-step explanation:

Given Y is the circumcenter of ΔSTU. we have to find the measures SV, SY, YW and YX.

As Circumcenter is equidistant from the vertices of triangle and also The circumcenter is the point at which the three perpendicular bisectors of the sides of the triangle meet.

Hence, VY, YW and YX are the perpendicular bisectors on the sides ST, TU and SU.

Given ST=18 units.

As VY is perpendicular bisector implies SV=9 units.

Also in triangle VTY

$YT^{2}=VY^{2}+VT^{2}$

⇒ $14^{2}=VY^{2}+9^{2}$

⇒ VY^{2}=115

As vertices of triangle are equidistant from the circumcenter

⇒ SY=YT=UY=14 units

Hence, SY is 14 units

In ΔUWY, $UY^{2}=YW^{2}+UW^{2}$

⇒ $14^{2}=YW^2+11^{2}$

⇒ $YW^2=196-121=75$ ⇒ YW= $\sqrt75units$

In ΔYXU, $UY^{2}=YX^{2}+XU^{2}$

⇒ $14^{2}=YX^2+13^{2}$

⇒ $YX^2=196-169=27$ ⇒ YW= $\sqrt27units$

Hence, measures are SV=9 units., SY=14 units, YW= $\sqrt75units$ , YW= $\sqrt27units$

4 0

2 years ago

The population of Smalltown in the year 1890 was 6,250. Since then, it has increased at a rate of 3.75% each year. • What was th

Vikki [24]

Answer:

year 1915 : 15,689

year 1940 : 39,381

t = 137.9 years

Step-by-step explanation:

Hi, to answer this question we have to apply an exponential growth function:

A = P (1 + r) t

Where:

p = original population

r = growing rate (decimal form)

t= years

A = population after t years

Replacing with the values given:

A = 6,250 (1 + 3.75/100)^t

A = 6,250 (1 + 0.0375)^t

A = 6,250 (1.0375)^t

So, by the year 1915 :

1915-1890 = 25 years passed (t)

A = 6,250 (1.0375)^25

A = 15,689

by the year 1940 :

1940-1890 = 50 years passed (t)

A = 6,250 (1.0375)^50

A = 39,381

When will the population reach 1,000,000?. We have to subtitute A=1000000 and solve for t.

1,000,000= 6,250 (1.0375)^t

1,000,000/ 6,250 =(1.0375)^t

160 = 1.0375^t

log 160 = log 1.0375^t

log 160 = (t ) log 1.0375

log160 / log 1.0375= t

t = 137.9 years

8 0

2 years ago