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vladimir2022 [97]

3 years ago

10

What are the integer solutions of the inequality |x| < 3? A. 0, 1, 2, and 3 B. 0, 1, and 2 C. –2, –1, 0, 1, and 2 D. –3, –2,

–1, 0, 1, 2, and 3

1 answer:

MaRussiya [10]3 years ago

4 0

The answer to your answer will be A

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If -1 and 2 are the two zeros of the polynomial x3-4x2-7x+10 find the third zero

rewona [7]

note that the zeroes should be x = 1 and x = - 2 not x = - 1 and x = 2

the factors are then ( x - 1 ) and (x + 2)

and x³ - 4x² - 7x + 10 = (x - 1 )(x + 2) = x² + x - 2

thus x² + x - 2 is a factor

dividing x³ - 4x² - 7x + 10 by x² + x - 2 gives x - 5

the third zero is x = 5

check : (5)³ -4(5)² - 7(5) + 10 = 125 - 100 - 35 + 10 = 0

5 0

3 years ago

Factorise = <br> 5m + 20

taurus [48]

So 20=2 itmes m2 times 5

5m=5 times m

so
5m=20
5 times m=2 times 2 times 5
divid eboth sides y 5
m=2 times 2
m=4

8 0

3 years ago

A bank in the Bay area is considering a training program for its staff. The probability that a new training program will increas

WITCHER [35]

Answer:

$P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68$

See explanation below.

Step-by-step explanation:

For this case we define first some notation:

A= A new training program will increase customer satisfaction ratings

B= The training program can be kept within the original budget allocation

And for these two events we have defined the following probabilities

$P(A) = 0.8, P(B) = 0.2$

We are assuming that the two events are independent so then we have the following propert:

$P(A \cap B ) = P(A) * P(B)$

And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:

$P(B' \cup A')$

And using the De Morgan laws we know that:

$(A \cap B)' = A' \cup B'$

So then we can write the probability like this:

$P(B' \cup A') = P((A \cap B)')$

And using the complement rule we can do this:

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

Since A and B are independent we have:

$P(A \cap B )=P(A)*P(B) =(0.8*0.4) =0.32$

And then our final answer would be:

$P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68$

5 0

3 years ago

Theres 96 seats and 12 tables I just need an equation.

Citrus2011 [14]

Answer:

s = 8t

Step-by-step explanation:

.............................

3 0

3 years ago

How many equivalence relations are there on the set 1, 2, 3]?

Alex787 [66]

Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

equivalence relation R on {1,2,3}

1.For reflexive, it must contain (1,1),(2,2),(3,3)

2.For transitive, it must satisfy: if (x,y)∈R then (y,x)∈R

3. For symmetric, it must satisfy: if (x,y)∈R,(y,z)∈R then (x,z)∈R

Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

This is because if (1,2) is in the relation then (2,1) must be there in the relation.

the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

we can have three possible equivalence relations:

{(1,1),(2,2),(3,3),(1,2),(2,1)}

{(1,1),(2,2),(3,3),(1,3),(3,1)}

{(1,1),(2,2),(3,3),(2,3),(3,2)}

6 0

3 years ago