All categories

3 years ago

What does it mean if an equation is true, false, or open? Give definitions of each.

1 answer:

6 0

I'm gonna do the example of this equation 2+2, as we all know when we add it together the sum of it is 4 , here are the examples: 2 + 2 = 4 is true
2 + 3 = 4 is false
x + 3 = 4 is open

You might be interested in

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

I need help with the last 2 I need the next 3 in the sequence and the nth term

kkurt [141]

Answer:

f. the sequence goes by the half of the previous number.

= 16, 8, 4, 2, 1, 1/2, 1/4, 1/8...

g. the sequence goes by adding the consecutive odd number added to the previous number.

first number= 3.

second number= 3+3

= 6

third number= 6+5

= 11

fourth number= 11+7

= 18

fifth number= 18+9

= 27

sixth number= 27+11

= 38

seventh number= 38+13

= 51, etc.

= 3, 6, 11, 18, 27, 38, 51...

7 0

3 years ago

A factory's worker productivity is normally distributed. one worker produces an average of 77 units per day with a standard devi

VLD [36.1K]

One worker produces an average of 84 units per day with a street What is the probability that in any single day worker 1 will outproduce worker 2? A) 0.1141.

Answer, factory worker productivity is normally distributed. One worker produces an average of 75 units per day with a standar, day with a standard deviation of 20. Another worker produces at an average rate of 65 per day.

4 0

3 years ago

If 3

Sliva [168]

Answer:Given that 3 persons can weave 168 shawls in 14 days. Then, Shawls are woven in 1 day by the 3 men = 168/14 = 12. Shawls are woven in 1 day ...

Step-by-step explanation:

4 0

2 years ago

How do i write ten more shoes than ruben as a variable

Lana71 [14]

Answer:

s + 10

Step-by-step explanation:

Ruben has s number of shoes. Then, 10 more shoes that Ruben has is

s + 10

7 0

3 years ago