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3 years ago

8

Solve graph and write the solution to each inequality in interval notation 1. |b+3|>7

1 answer:

OleMash [197]3 years ago

3 0

B+3> 7 and. b+3< -7
-3 -3 -3. -3

b>4 and b<-10

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A website posts evaluations on airlines. It posts the most positive and the most negative comments for each airline. Discuss the

Fittoniya [83]

Answer:

B. the reader cannot accurately assess the airline because the evaluations are not necessarily representative of all evaluations.

D. the reader cannot accurately assess the airline without reading all the evaluations.

Step-by-step explanation:

Using the most positive and the most negative comments to evaluate the airlines will not be an accurate measure of the airline because these are not a good enough sample. The opinions of those who fall in-between the most positive and negative should also be included as they could have some vital details as well.

Also, in order to be completely accurate, all the comments of the airline should have been posted because this would represent the entire population of comments and so would give the best assessment.

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2 years ago

Find the product of - 1.3 and 2.12 ?

Alchen [17]

The answer is -2.756

8 0

2 years ago

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A label that wraps around a box of golf balls covers 75% of its lateral surface area. What is the value of x? The height is 2 th

GalinKa [24]

To find the area of the rectangular prism, use the formula:

A = width x height x length

A = 2 x 2 x 3

A = 12

Now multiply A with 0.75

12 x 0.75 = 9

x = 9 in³

hope this helps

5 0

3 years ago

Read 2 more answers

The independent variable in the relationship is the and should be placed on the . The dependent variable in the relationship is

Gelneren [198K]

Answer: The independent variable is what your are testing.

The dependent variable is your outcome is.

Step-by-step explanation:

3 0

2 years ago

Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

4 0

2 years ago