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Mice21 [21]
3 years ago
7

A particle is moving on a straight line with velocity

Mathematics
1 answer:
Lena [83]3 years ago
7 0

v =  {t}^{2}  - 6t + 5

t=2

v =  {2}^{2}  - 6 \times 2 + 5 =  - 3

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A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
AURORKA [14]

Answer:

(−0.103371 ; 0.063371) ;

No ;

( -0.0463642, 0.0063642)

Step-by-step explanation:

Shift 1:

Sample size, n1 = 30

Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm

Shift 2:

Sample size, n2 = 25

Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17

Mean difference ; μ1 - μ2

Zcritical at 95% confidence interval = 1.96

Using the relation :

(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)

(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)

Lower boundary :

-0.02 - 0.0833710 = −0.103371

Upper boundary :

-0.02 + 0.0833710 = 0.063371

(−0.103371 ; 0.063371)

B.)

We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.

C.)

For sample size :

n1 = 300 ; n2 = 250

(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)

Lower boundary :

-0.02 - 0.0263642 = −0.0463642

Upper boundary :

-0.02 + 0.0263642 = 0.0063642

( -0.0463642, 0.0063642)

7 0
2 years ago
Multiply.
KATRIN_1 [288]
The answer could be
-1287
im not sure
7 0
3 years ago
A lottery consists of one $2000 winner, three $500 winners, and ten $100 winners. a total of 1000 tickets are sold for $10 each.
Semenov [28]

Answer: -5.5

 

WOKINGS

Given that the lottery has the following number of winners:

One $2000 winner

Three $500 winners

Ten $100 winners

 

Also Given,

A total of 1000 tickets are sold

Each ticket costs $10

 

The expected winning for a person purchasing one ticket is the sum of the products of the gain/loss and their corresponding probability.

 

There is one $2000 winner

There are 1000 tickets

The probability of winning $2000

= 1/1000

= .001

There are three $500 winners

There are 1000 tickets

The probability of winning $500

= 3/1000

= .003

 


There are ten $100 winners

There are 1000 tickets

The probability of winning $100

= 10/1000

= .01

 

Since each ticket costs $10

Everyone who buys a ticket automatically loses $10.

Therefore, the probability of losing $10 is 1

 

Now to calculate the expected winning for a person purchasing one ticket

= 2000(.001) + 500(.003) + 100(.01) – 10(1)

= 2 + 1.5 + 1 – 10

= -5.5

 

The expected winning is -5.5. This implies that a person playing this lottery can expect to lose $5.50 for every one ticket that they purchase. 

7 0
3 years ago
Emil bought a camera for $268.26, including tax. He made a down payment of $12.00 and paid the balance in 6 equal monthly paymen
iragen [17]
The total cost of the camera is $268.26

A down payment of $12 was paid so he now owes $256.26

Every 6 months Emil has to pay:
256.26 / 6 = 42.71
Therefore, each moth, Emil has to pay $42.71 
7 0
3 years ago
Read 2 more answers
A student takes an exam containing 1818 true or false questions. If the student guesses, what is the probability that he will ge
creativ13 [48]

Probability

p=66×100/1818=3.6303630363%

After rounding to 4 decimals:

p=3.6304%

8 0
2 years ago
Read 2 more answers
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