All categories

3 years ago

What is negative twenty times three? −20 ⋅ 3

Mathematics

2 answers:

bulgar [2K]3 years ago

7 0

Negative × positive = negative
20×3=60
since its a negative, just add the sign
-60

neonofarm [45]3 years ago

4 0

-60 because a negative times a positive is always negative

You might be interested in

What’s the answer to this question?

denis23 [38]

I think the answer is C ( 11 )

8 0

3 years ago

what is the coordinate of point A after being rotated 90° clockwise, if the coordinates to begin with are (-8,-1)

lys-0071 [83]

...-1,8......................................

8 0

4 years ago

About 13% of the population of a large country is nervous around strangersnervous around strangers. If two people are randomly

Molodets [167]

Answer:

1.69% probability both are nervous around strangers.

24.31% probability at least one is nervous around strangers.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are nervous around strangers, or they are not. The people are chosen randomly, which means that the probability of a person being nervous around strangers is independent from other people. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

About 13% of the population of a large country is nervous around strangers.

This means that $p = 0.13$ .

Two people are randomly selected

This means that $n = 2$ .

What is the probability both are nervous around strangers?

This is $P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.13)^{2}.(0.87)^{0} = 0.0169$

1.69% probability both are nervous around strangers.

What is the probability at least one is nervous around strangersnervous around strangers?

This is $P(X \geq 1)$

$P(X \geq 1) = P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{2,1}.(0.13)^{1}.(0.87)^{1} = 0.2262$

$P(X = 2) = C_{2,2}.(0.13)^{2}.(0.87)^{0} = 0.0169$

$P(X \geq 1) = P(X = 1) + P(X = 2) = 0.2262 + 0.0169 = 0.2431$

24.31% probability at least one is nervous around strangers.

3 0

3 years ago

Drag each expression to the correct location on the table.

ololo11 [35]

Answer:

see the picture

Step-by-step explanation:

attached

4 0

3 years ago

An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the A

Virty [35]

Answer:

0.5

option A

0.6

0.1

Step-by-step explanation:

The event A and B are independent so

P(A∩B)=P(A)*P(B)

P(A∩B)=P(0.2)*P(0.5)=0.10

We have to find P(B'|A')

P(B'|A')=P(B'∩A')/P(A')

P(A)=0.2

P(A')=Asian project is not successful=1-P(A)=1-0.2=0.8

P(B)=0.5

P(B')=Europe project is not successful=1-P(B)=1-0.5=0.5

P(B'∩A')=Europe and Asia both project are not successful=P(A')*P(B')=0.8*0.5=0.4

P(B'|A')=P(B'∩A')/P(A')=0.4/0.8=0.5

This can be done by another independence property for conditional probability

P(B|A)=P(B)

P(B'|A')=P(B')

P(B'|A')=0.5

Probability of at least one of two projects will be successful means that the probability of success of Asia project or probability of success of Europe project or probability of success of Europe and Asian project which is P(AUB).

P(AUB)=P(A)+P(B)-P(A∩B)

P(AUB)=0.2+0.5-0.1

P(AUB)=0.6

Probability of only Asian project is successful given that at least one of the two projects is successful means that probability of success of project Asia while the project Europe is not successful denoted as P((A∩B')/(A∪B))=?

P((A∩B')/(A∪B))=P((A∩B')∩(A∪B))/P(A∪B)

P((A∩B')∩(A∪B))=P(A∩B')*P(A∪B)

P(A∩B')=P(A)*P(B')=0.2*0.5=0.10

P((A∩B')∩(A∪B))=0.1*0.6=0.06

P((A∩B')/(A∪B))=0.06/0.6=0.1

4 0

3 years ago