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laiz [17]
1 year ago
8

PLEASE HELP! The avenues in a particular city run north to south and are numbered consecutively with 1st Avenue at the western b

order of the city. The streets in the city run east to west and are numbered consecutively with 1st Street at the southern border of the city. For a festival, the city is not allowing cars to park in a rectangular region bordered by 6th Avenue to the west, 10th Avenue to the east, 11th Street to the south, and 13th Street to the north. If x is the
avenue number and yis the street number, which of the following systems describes the region in which cars are not allowed to park?

Mathematics
2 answers:
AveGali [126]1 year ago
4 0

Answer: D

Step-by-step explanation:

Just did it

maksim [4K]1 year ago
3 0

Answer:

D.

Step-by-step explanation:

Just did it on a.p.e.x,

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Jane is taking two courses. The probablity she passes the first course is 0,7. The probablity she passes the second course is 0.
makkiz [27]

Answer:

a. 0.6

b. not independent

c. 0.1

d. 0.4

e. 0.3

Step-by-step explanation:

a.

P(passing first course)=P(C1)=0.7

P(passing second course)=P(C2)=0.8

P(passing at least one course)=P(C1∪C2)=0.9

P( passes both courses)=P(C1∩C2)=?

We know that

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∩B)=P(A)+P(B)-P(A∪B)

So,

P( passes both courses)=P(C1∩C2)=P(C1)+P(C2)-P(C1∪C2)

P( passes both courses)=P(C1∩C2)=0.7+0.8-0.9

P( passes both courses)=P(C1∩C2)=0.6

Thus, the probability she passes both courses is 0.6.

b.

The event of passing one course is independent of passing another course if

P(C1∩C2)=P(C1)*P(C2)

P(C1)*P(C2)=0.7*0.8=0.56

P(C1∩C2)=0.6

As,

0.6≠0.56

P(C1∩C2)≠P(C1)*P(C2),

So, the event of passing one course is dependent of passing another course.

c.

P(not passing either course)=P(C1∪C2)'=1-P(C1∪C2)

P(not passing either course)=P(C1∪C2)'=1-0.9

P(not passing either course)=P(C1∪C2)'=0.1

Thus, the probability of not passing either course is 0.1.

d.

P(not passing both courses)=P(C1∩C2)'=1-P(C1∩C2)

P(not passing both courses)=P(C1∩C2)'=1-0.6

P(not passing both courses)=P(C1∩C2)'=0.4

Thus, the probability of not passing both courses is 0.4.

e.

P(passing exactly one course)=?

P(passing exactly course 1)=P(C1)-P(C1∩C2)=0.7-0.6=0.1

P(passing exactly course 2)=P(C2)-P(C1∩C2)=0.8-0.6=0.2

P(passing exactly one course)=P(passing exactly course 1)+P(passing exactly course 2)

P(passing exactly one course)=0.1+0.2

P(passing exactly one course)=0.3

Thus, the probability of passing exactly one course is 0.3.

3 0
1 year ago
Liam volunteers to help plan the event. The first month, he volunteered for x hours. The next month, he volunteered for 1 2/3 ti
Eddi Din [679]

Answer:

<u>Hours in first month = 8</u>

Step-by-step explanation:

AS it is given

He volunteered fr the first month = x hours

He volunteered for the second  month = 12/3 times of first month

which will be = 12/3 * x

Total Hours volunteered = 40 hours

Now according to the given conditions

Volunteered first month + volunteered second month = total

putting this gives

x + 12/3 8 x  = 40

as 12/3 = 4 so

it becomes

x + 4x = 40    

solcing it will gives

5x = 40

x = 40/5

x = 8

<u><em>So Numbers of hours worked in first month are 8</em></u>

<u><em></em></u>

<u>I hope this help you! :)</u>

6 0
1 year ago
EAB and DCB are two right triangles. The figure has BED≅ BDE. Point B is the midpoint of segment AC. Prove: EAB≅ DCB
aleksklad [387]
See the attached picture. 

<span>you are given that ABCE is an isosceles trapezoid. </span>

<span>you are given that AB is parallel to EC. </span>

<span>this means that AE is congruent to BC. </span>

<span>you are given that AE and AD are congruent. </span>

<span>triangle EAD is an isosceles triangle because AE and AD are congruent. </span>

<span>this means that angle 1 is equal to angle 3. </span>

<span>since angle 1 is equal to angle 2 and angle 3 is equal to angle 1, then angle 3 is also equal to angle 2. </span>

<span>this means that AD and BC are parellel because their corresponding angles (angles 3 and 2) are equal. </span>

<span>since AB is parallel to EC and DC is part of the same line, than AB is parallel to DC. </span>

<span>you have AB parallel to DC and AD parallel to BC. </span>

<span>if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. </span>

<span>that might be able to do it,depending on whether all these statements are acceptable without proof. </span>

<span>they are either postulates or theorems that have been previously proven. </span>

<span>if not, then you need to go a little deeper and prove some of the statements that you used.. </span>

here's my diagram. 

 

<span>this is not a formal proof, but should give you some ideas about how to proceed. </span>

<span>you can also prove that angle 4 is equal to angle 2 because they are alternate interior angles of parallel lines. </span>

<span>you can also prove that angle 6 is equal to angle 5 because they are alternate interior angles of parallel lines. </span>
5 0
1 year ago
Cody works at a car dealership where he earns $300 each week plus a commission equal to 1% of his total sales. This week his goa
GarryVolchara [31]

Answer:

Step-by-step explanation:

Hepl me wath is the answer

5 0
2 years ago
7 divided by 423 full answer
anygoal [31]
The answer is 0.01654846336
3 0
1 year ago
Read 2 more answers
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