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3 years ago

166. Запишите все целые числа, расположенные на координатной прямой между числами:

Mathematics

1 answer:

FromTheMoon [43]3 years ago

6 0

Answer:

Don’t speak that language sorry maybe translate into English?

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PLEASE HELP

Nastasia [14]

Only if i knew :(((:(((

5 0

3 years ago

If you apply the changes below to the quadratic parent function, f(x)=x^2 which of these is the equation of the new function?

g100num [7]

Answer:

Step-by-step explanation:

Given the parent function y = x², then

y = (x + a) represents a horizontal translation

• If a > 0 , then shift to the left of a units

• If a < 0, then shift to the right of a units

Thus a shift of 3 units left ⇒ y = (x + 3)²

------------------------------------------------------------

A vertical stretch of a units, stretches away from the x- axis

A vertical compression of a units, compresses towards the x- axis

• | a | > 1 is a stretch

• 0 < | a | < 1 is a compression

Thus a vertical stretch of 4 ⇒ y = 4(x + 3)²

--------------------------------------------------------------

A reflection in the x- axis, negates the y- coordinate

y = - 4(x + 3)² → A

5 0

4 years ago

Reciprocal of 12 by 15 is

romanna [79]

the reciprocal is 15 by 12

5 0

3 years ago

Determine the quadrant(s) in which (x,y) is located so that the condition(s) is (are) satisfied.

polet [3.4K]

Quadrant 2 is the only one that can satisfy the condition. In there, x<0 and y>0

7 0

3 years ago

Suppose that a recent poll of American households about car ownership found that for households with a car, 39% owned a sedan, 3

Rama09 [41]

Answer:

The probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.

Step-by-step explanation:

Let X = number of household owns a sports car.

The probability of X is, P (X) = p = 0.07.

Then the random variable X follows a Binomial distribution with n = 3 and p = 0.07.

The probability function of a binomial distribution is:

$P(X=x) = {n\choose x}p^{x}[1-p]^{n-x}\\$

Compute the probability that of the 3 households randomly selected at least 1 owns a sports car:

$P(X\geq 1)=1-P(X$

Thus, the probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.

4 0

3 years ago