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katrin [286]
3 years ago
10

2010N of force are required to move a 16kg object. How fast is the Object accelerating?

Mathematics
1 answer:
Blizzard [7]3 years ago
6 0
Acceleration = mass / force

16 kg / 2010 N = 0.007960199004975 m/s squared
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Solve the system of equations by substitution.<br> y = 5x - 13<br> 4x + 3y = 18<br> The solution is
Rudik [331]

Answer:

x=3 and y=2

Step-by-step explanation:

1.) 4x+3(5x-13)=18

expand into 4x+15x-39=18

18+39= 57

15x+4x= 19x

57/19= 3 so x=3

substitute 3 into y= 5x - 13

and you get 2 so y=2

3 0
3 years ago
What’s is the gcf 33c,55cd
Juli2301 [7.4K]
The GCF of 33c and 55cd is

33c/11c = 3
55cd/11c = 5d

the greatest common factor is 11c 

hope this helps
4 0
3 years ago
Which of the following is(are) true? I. The mean of a population depends on the particular sample chosen. II. The standard devia
Margarita [4]

Answer:

II and III

Step-by-step explanation:

From statement II in the question, it is true that the standard deviations of two different samples from the same population may be the same. The population standard deviation is a fixed value calculated from every individual in the population.  A sample standard deviation is calculated from only some of the individuals in a population.

Also from statement III, it is true that statistical inferences can be used to draw conclusions about the populations based on sample data. The mean of a population does not necessarily depends on the particular sample chosen. Therefore statement I is false.  

3 0
3 years ago
Jay took out a small-business loan for 210,000. The terms of the loan are 5% annual interest for four years.
Marta_Voda [28]
5% of 210000=
(210000/100)5= $10500 anually

so 4 years will be $42000

he'll pay 42000 + 210000
$252000
answer is B

4 0
3 years ago
Read 2 more answers
Consider a Poisson distribution with μ = 6.
bearhunter [10]

Answer:

a) P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b) f(2) = 0.04462

c) f(1) = 0.01487

d) P(X \geq 2) = 0.93803

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of successes

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

In this question:

\mu = 6

a. Write the appropriate Poisson probability function.

Considering \mu = 6

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b. Compute f (2).

This is P(X = 2). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

P(X \geq 2) = 1 - P(X < 2)

In which:

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

Then

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803

So

P(X \geq 2) = 0.93803

5 0
2 years ago
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