1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bagirrra123 [75]
4 years ago
11

Last question!!!!!!! Is Number 10 correct?

Mathematics
1 answer:
natali 33 [55]4 years ago
3 0
Number 10 for what? i don't see anything
You might be interested in
(4a2 - 3b2)(16a4 +12a2b2 + 9b4)
Lynna [10]

Answer: the simplify anwser is 64a^ 6 - 27b^6

Step-by-step explanation:

5 0
3 years ago
Show the original equation for x
Molodets [167]

Answer:

whats the equation?

Step-by-step explanation:

3 0
3 years ago
$7.50 admission; 20% off 5.75% tax
yuradex [85]

hey I just met you and this is crazy but here's my number so call me baby

8 0
3 years ago
Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual
Stells [14]

Answer:

(a) P( A ∩ B )=0.5 is not possible.

(b) 0.7

(c) 0.3

(d) 0.3

(e) 0.4

Step-by-step explanation:

Given information: The alphabet A and B represents the following events

A : Individual has a Visa credit card.

B: Individual has a MasterCard.

P(A)= 0.6 and P(B)=0.4.

(a)

We need to check whether P( A ∩ B ) can be 0.5 or not.

A\cap B\subset A and A\cap B\subset B

P(A\cap B)\leq P(A) and P(A\cap B)\leq P(B)

P(A\cap B)\leq 0.6 and P(A\cap B)\leq 0.4

From these two inequalities we conclude that

P(A\cap B)\leq 0.4

Therefore, P( A ∩ B )=0.5 is not possible.

(b)

Let P(A\cap B)=0.3

We need to find the probability that student has one of these two types of cards.

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Substitute the given values.

P(A\cup B)=0.6+0.4-0.3=0.7

Therefore the probability that student has one of these two types of cards is 0.7.

(c)

We need to find the probability that the selected student has neither type of card.

P(A'\cup B')=1-P(A\cup B)

P(A'\cup B')=1-0.7=0.3

Therefore the probability that the selected student has neither type of card is 0.3.

(d)

The event that the select student has a visa card, but not a mastercard is defined as

A-B

It can also written as

A\cap B'

The probability of this event is

P(A\cap B')=P(A)-P(A\cap B)

P(A\cap B')=0.6-0.3=0.3

Therefore the probability that the select student has a visa card, but not a mastercard is 0.3.

(e)

We need to find the probability that the selected student has exactly one of the two types of cards.

P(A\cap B')+P(A\cap B')=P(A\cup B)-P(A\cap B)

P(A\cap B')+P(A\cap B')=0.7-0.3

P(A\cap B')+P(A\cap B')=0.4

Therefore the probability that the selected student has exactly one of the two types of cards is 0.4.

8 0
3 years ago
What is 8percent of 575
QveST [7]

Answer:

46%hope this helps add me

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
Other questions:
  • Does anyone know the answer ?
    15·1 answer
  • Which expression represents 7 times the difference between 300 and 28
    6·2 answers
  • Choose one of the factors of 1331x3 − 8y3.
    11·2 answers
  • 30 points. Precalculus HELP.
    8·1 answer
  • What is 76% of $25,000?
    14·2 answers
  • Solve for n<br> (1.0425)^4t=3
    5·2 answers
  • True or False?
    15·2 answers
  • Find the measure of the arc or angle indicated.
    11·1 answer
  • The Point ( 5 , − 2 ) is translated to the left 3 units. What are the new coordinates? (8, −2) (8, −2) (5, 1) (5, 1) (5, −5 (5,
    6·1 answer
  • Erin has a rope that is 3 yards long.
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!