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3 years ago

Find the value of x.

Mathematics

1 answer:

Anastasy [175]3 years ago

8 0

Answer:

Step-by-step explanation:

Because they are the same

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Chords AB and CD intersect at point M. According to the diagram, which pair of numbers could represent lengths of segments CM

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Answer:

Answer:

Step-by-step explanation:

Since the product of the intersecting chords should be equal

==> AM x BM = CM x DM

==> 4 x 5 = CM x DM

==> 20 = CM x DM

Therefore CM x DM = 2 X 10

Step-by-step explanation:

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What is the slope in the equations: y = 4x + 3

Anestetic [448]

Using the slope-intercept form, the slope is 4

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I opened a grocery account at my bank with $250. Every week thereafter, I withdraw $90 from the account to pay for groceries. If

Ksenya-84 [330]

Answer:

y= -90x+250

Step-by-step explanation:

From the information given, the equation would be that the amount in the account would be equal to -90 that is the amount you withdraw from the account each week multiply for x that is the number of weeks plus $250 that is your initial balance:

y= -90x+250, where

y= amount in the account

x= number of weeks.

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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

(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.

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What are two more ways to arrange the cars into even rows?

kotykmax [81]

Answer:

make both sides even by putting the same amount of cars on either side

Step-by-step explanation:

My pretty face is the explanation on how smart I am lol

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