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3 years ago

Find the common property and use it to describe these sets. C={1,4,9,16} C is a set of x, where x depends on n.

Mathematics

2 answers:

Korolek [52]3 years ago

8 0

Answer:

x depends on natural number.

PilotLPTM [1.2K]3 years ago

6 0

Answer:

C={1,4,9,16} C is a set of x, where x depends on .

Step-by-step explanation:

x depends on natural number of.

hope it is helpful to you

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A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

erica [24]

Answer:

(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

Let $E_1, E_2,$ and $E_3$ be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

where $P(E_2/E_1)$ denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

$P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)$

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

$P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)$

$\Rightarrow P(F)=(1-0.9)+(1-0.8)+(1-0.7)=0.6$

Lef $F_2$ be the event that she failed the 2nd exam, so

$P(F_2)=1-\frac{P(E_2)}{P(E_1)}$

$\Rightarrow P(F_2)=1-0.8=0.2$

So, the conditional probability that she failed the 2nd exam is

$P\left(\frac{F_2}{F}\right)=\frac{P(F_2)}{P(F)}$

$\Rightarrow P\left(\frc{F_2}{F}\right)=\frac{0.2}{0.6}=\frac{1}{3}=0.33$

5 0

3 years ago

PLEASE HELP: A standard six-sided die is rolled.

amm1812

Answer in fraction form is 1/3

Answer in decimal form is 0.3333

Pick one answer only.

==========================================================

Explanation:

The sample space is the set of all possible outcomes. In this case, the outcomes consist of values between 1 and 6

S = sample space

S = {1,2,3,4,5,6}

There are 6 items here. Let B = 6.

We want to roll a number greater than 4, so the event space we're after is

E = {5,6}

which consists of 2 items. Let A = 2.

The probability we want is A/B = 2/6 = 1/3 = 0.3333

So if you go with the fraction option, then you'll type in 1/3

If you go with the decimal option, then you'll type in 0.3333

4 0

3 years ago

(3x + 3x²) + (2x² + 2x)?

Galina-37 [17]

Answer:

=5x2+5x

Step-by-step explanation:

Let's simplify step-by-step.

3x+3x2+2x2+2x

8 0

3 years ago

A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome

Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable X represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), p = 0.52

The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.

(1)

Compute the probability of overbooking as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

$=P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135$

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

$=P(X$

Thus, the probability that the flight has empty seats is 0.4625.

4 0

3 years ago

The length of the diagonal of a square is 30square root 2 find the perimeter of the square

RoseWind [281]

Answer:

120.

Step-by-step explanation:

Using the Pythagoras theorem:

(30√2)^2 = x^2 + x^2 where x =length of each side of the square

1800 = 2x^2

x^2 = 900

x = 30.

So the perimeter = 4*30 = 120.

5 0

3 years ago