Expand the following:
(x - 6) (3 x^2 + 10 x - 1)
Hint: | Multiply out (x - 6) (3 x^2 + 10 x - 1).
| | | | x | - | 6
| | 3 x^2 | + | 10 x | - | 1
| | | | -x | + | 6
| | 10 x^2 | - | 60 x | + | 0
3 x^3 | - | 18 x^2 | + | 0 x | + | 0
3 x^3 | - | 8 x^2 | - | 61 x | + | 6:
Answer: 3 x^3 - 8 x^2 - 61 x + 6 Thus B:

7 0

3 years ago

Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected a

charle [14.2K]

Answer:

Step-by-step explanation:

Confidence interval in less than symbol expressed as

$\bar{x} - E < \mu < \bar{x} + E
$

Where $\bar{x}$ is sample mean and $E$ is margin of error.

$13.03 < \mu 13.42
$

The given t interval is $(13.032 , 13.418 )
$

That is $\bar{x} - E = 13.032$ and $\bar{x} + E = 13.418$

Solve these two equation by adding together.

$2 \bar{x} = 13.032 + 13.418
\\\\\bar{x} = 13.225$

Solve this value of \bar{x} in equation $\bar{x} - E = 13.032$ and solve for $E$

$13.225 - E = 13.032
\\\\E = 0.193
$

Best point estimate of $\mu = \bar{x} = 13.225
$

Best point estimate of margin of error = 0.193

Since sample size = 100 which is sufficiently large (Greater than 30) , it is no need to confirm that

sample data appear to be form a population with normal distribution.

8 0

3 years ago

Describe the rotation of J

Serga [27]

Answer:

The coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1)

The coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)

Step-by-step explanation:

Square JKLM with vertices

J(1, -3)

K(5, 0)

L(8, -4)

M(4, -7)

We have to determine the answer for the image J' of the point (1, -3) when we rotate the point by 90° counterclockwise, we need to switch x and y, make y negative.

In other words, the rule to rotate a point by 90° counterclockwise.

P(x, y) → P'(-y, x)

As we are given that J(1, -3), so the coordinates of J' will be:

J(1, -3) → J'(3, 1)

Therefore, the coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1).

When the point is rotated by 90° clockwise, we need to switch x and y, make x negative.

In other words, the rule to rotate a point by 90° clockwise.

P(x, y) → P'(y, -x)

As we are given that J(1, -3), so the coordinates of J' will be:

J(1, -3) → J'(-3, -1)

Therefore, the coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)

3 0

3 years ago

1 2 4 3 6 8 7 14 16? whats the next term in this sequence?

ANEK [815]

The next term in this sequence would be 18.

8 0

3 years ago