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2 years ago

Complete the table for the equation y = x 2. Then use the table to graph the equation.

Mathematics

2 answers:

Vladimir79 [104]2 years ago

7 0

Yes 101010101.quagga

Brrunno [24]2 years ago

3 0

The coordinates (starting from left) are
(-2,-4)(-1,-3)(0,-2)(1,-1)(2,0)

to find the y..
- ur given y=x-2
- u have all the x values on the table

so say u wanna find the y for -2, just plug it into the equation y = x - 2
so then it’ll be y = (-2) - 2
the only reason u can plug in the -2 is bc the -2 is a x value

same logic for the other numbers on the table, to find the y value for -1 just plug in -1 into y = x - 2
so then it’ll be y = (-1) - 2

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What kind of sequence is the pattern 1, 6, 7, 13, 20, ...?

AlexFokin [52]

Answer:

The given sequence 6, 7, 13, 20, ... is a recursive sequence

Step-by-step explanation:

As the given sequence is

$6, 7, 13, 20, ...$

It cannot be an arithmetic sequence as the common difference between two consecutive terms in not constant.

$d = 7-6=1$ , $d = 13-7=6$

As d is not same. Hence, it cannot be an arithmetic sequence.

It also cannot be a geometrical sequence and exponential sequence.

It cannot be geometric sequence as the common ratio between two consecutive terms in not constant.

$r = 7-6=1$ , $r = 13-7=6$

$r = \frac{7}{6}$ , $r = \frac{13}{7}$

As r is not same, Hence, it cannot be a geometric sequence or exponential sequence. As exponential sequence and geometric sequence are basically the same thing.

So, if we carefully observe, we can determine that:

The given sequence 6, 7, 13, 20, ... is a recursive sequence.

Please have a close look that each term is being created by adding the preceding two terms.

For example, the sequence is generated by starting from 1.

${\displaystyle F_{1}=1$

and

$F_{n}=F_{n-1}+F_{n-2}$

for n > 1.

Keywords: sequence, arithmetic sequence, geometric sequence, exponential sequence

Learn more about sequence from brainly.com/question/10986621

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6 0

3 years ago

At a university, 60% of the students who apply. Of the students the university accepts, 25% actually enroll If 2000 students

sweet-ann [11.9K]

Answer:

300 students are enrolled.

Step-by-step explanation:

As the university accepts 60% of the students who apply.

As 2000 students apply, it means that:

2000 × 60% ⇒ 2000 × 0.6 = 1200 students are accepted

Out of these 1200, 25% enrolled

1200 × 25% ⇒ 1200 × 0.25 = 300 students are enrolled

Therefore, 300 students are enrolled.

5 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$