All categories

3 years ago

Can someone please help me it’s due in 5 minutes

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

4 0

Answer:

You can use Desmos to graph any equation:

You might be interested in

Hi Guys

BabaBlast [244]

Answer:

x^2-2x+5x-10

= x^2+3x-10

Hope it helps.

6 0

4 years ago

Read 2 more answers

Stacy buys a bagel assortment that has 6 oat bagels for every 2 blueberry bagels. Naomi buys a bagel assortment that has 9 oat b

sdas [7]

Answer: Stacy: 30 Naomi: 18

Step-by-step explanation:

Stacy- ratio of oat to blueberry is 6:2 we want to find x:10

10 is 5 times 2, and whatever you do to the bottom you have to do to the top, and 6 times 5 is 30

Naomi- ratio of oat to blueberry is 9:5 we want to find x:10

The 10 is 2 times 5, and whatever you do to the bottom you have to do to the top, and 9 times 2 is 18

3 0

2 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

(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

Please help! I will give Brainliest!<br> Who is correct?

AlekseyPX

Answer:

Eric is correct; side must be between 4 and 16 inches

Step-by-step explanation:

triangle inequality theorem:

6 + 10 > x; 16 > x

x + 6 > 10; x > 4

4 < x < 16

8 0

3 years ago

Which statement is true about the graphs of the two lines y = –6 and x = 1/6 ?

Novay_Z [31]

They both intersect eachother

4 0

4 years ago

Read 2 more answers