F(X) = -X +2:Find f(X- 4)

The standard deviation of a sample taken from population A is 17.6 for a sample of 25. The standard deviation of a sample taken

agasfer [191]

33.3..........................................

3 0

3 years ago

a fox a coyote and a wolf weigh a total of 120 pounds the ratio of the weight of fox to the weight of coyote is 3:8 the weight o

ira [324]

There is not enough information to calculate this.

Knowing the weight ratio of the fox to coyote as 3:8 in no way allows you to know the respective ratio of the wolf. To know the weight of the wolf would require knowing its ratio value, then the weights of all 3 is an easy calculation. 
Example - 3:8:15 (f:c:w) is a plausible ratio based upon real-world weight averages for certain species/subspecies of the three. 
- knowing the values of the 3 terms as 3:8:15 gives a total of 3+8+15 = 26 ratio values 
- you then simply divide the total weight by this ratio value total; 120/26 = 4.62 
- so each ratio value is 4.62 units of weight*** 
- now, simply calculate the weight of each canid by multiplying its ratio value by the unit of weight... 
fox; 3 x 4.62 = 13.86 
coyote; 8 x 4.62 = 36.96 
wolf; 15 x 4.62 = 69.3 
Validate the ratios by adding the weights together (we should get 120) 13.86 + 36.96 + 69.3 = 120.12 
The total is slightly out because that 4.62 figure was a rounding up. 

Now, the thing is, there is nothing given that allows us to know exactly what ratio value the wolf should be, I chose 15 myself because that is a real-world plausible value when compared to 3:8 for the other 2. Changing it to 16, say, means that there are now 27 ratio values total giving a ratio value of 120/27 = 4.44 obviously changing the weights of all 3.

5 0

3 years ago

There are 10 less red skittles than orange skittles in the bag. The orange skittles are also twice the number of red skittles. F

Dmitry_Shevchenko [17]

20 skittles double the number of red

the red skittles have 10 skittles the orange skittles have 20 so its additon

20

+

10

------

30

6 0

3 years ago

Read 2 more answers

algebraically determine the values of h and k to correctly complete the identity stated below : 2 x cubed - 10x squared + 11x -

11Alexandr11 [23.1K]

Answer:

The values for expression is h = - 2 and k = 5

Step-by-step explanation:

Given algebraic expression can be written as :

2 x³ - 10 x² + 11 x - 7 = ( x - 4 ) × ( 2 x² + h x + 3 ) + k

Now opening the bracket

Or, 2 x³ - 10 x² + 11 x - 7 = x × ( 2 x² + h x + 3 ) - 4 × ( 2 x² + h x + 3 ) + k

Or, 2 x³ - 10 x² + 11 x - 7 = 2 x³ + h x² + 3 x - 2 x² - 4 h x - 12 +k

Or , 2 x³ - 10 x² + 11 x - 7 = 2 x³ + ( h - 2 ) x² + ( 3 - 4 h ) x - 12 + k

Now, equating the equation both sides

I.e - 10 = ( h - 2 )

Or , h - 2 = - 10

I.e , h = - 10 + 2

∴ h = - 2

Again , 11 = ( 3 - 4 h )

or, 11 = 3 - 4 h

or, 11 - 3 = - 4 h

or, 8 = - 4 h

∴ h = $\frac{-8}{4}$

I.e h = - 2

Again

- 7 = - 12 + k

Or, k = - 7 + 12

∴ k = 5

Hence The values for expression is h = - 2 and k = 5 . Answer

5 0

2 years ago

A sequence of Bernoulli trials consists of choosing components at random from a batch of components. A selected component is eit

Drupady [299]

Answer:

a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

Step-by-step explanation:

A sequence of Bernoulli trials composes the binomial distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that a selected component is non-defective is 0.8

This means that $p = 0.8$

a) Three non-defective components in a batch of seven components.

This is P(X = 3) when n = 7. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{7,3}.(0.8)^{3}.(0.2)^{4} = 0.0287$

0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 8 non-defective components are drawn before the first defective component is chosen.

Now the order is important, so the we just multiply the probabilities.

8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So

$p = (0.8)^8*0.2 = 0.0336$

0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

4 0

3 years ago