Ask question

Ask question

All categories

4 years ago

10

This graph shows the US unemployment rate from August 2010 to November 2011.

1 answer:

sergeinik [125]4 years ago

6 0

Answer: Will continue to rise

Step-by-step explanation:

Looking at the graph one notices that after a slight dip in the unemployment rate from August 2010 to January 2011, the unemployment rate began to rise and by November 2011 was still rising.

The arrow on the graph serves to indicate the direction the unemployment rate is going and as it is pointing upwards, this means that the Unemployment rate will continue to rise.

This was down to the fact that in 2011 the US was still yet to recover from the Great Recession of 2008 - 2009.

You might be interested in

Find the measure of

Bumek [7]

Answer:

80 I THINK, SORRY IF NOT RIGHT

Step-by-step explanation:

5 0

3 years ago

The population of a small town on January 1, 2005, was 7,200. On January 1, 2006, the population was 7,416. On January 1, 2007,

gogolik [260]

Answer:

(c) 8597

Step-by-step explanation:

A tool for regression analysis shows the given population values fit an exponential curve better than they fit a line. The exponential function for the population can be written as approximately ...

p(x) = 6211.54(1.02998^x) . . . . . x is years after 2000

Then for x=11, the population is about ...

p(11) ≈ 8596.4

The closest answer choice is 8597.

5 0

2 years ago

Calculate the double integral. $$\iint_{R}{\color{red}4} xye^{x^{2}y}\hspace*{3pt}dA, \quad R = [0, 1] \times [0, {\color{red}7}

Lady_Fox [76]

Answer:

$\mathbf{\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 (e^7 -8)}$

Step-by-step explanation:

Given that:

$\int \int _R 4xye^{x^2 \ y} \ dA, R = [0,1]\times [0,7]$

The rectangle R = [0,1] × [0,7]

R = { (x,y): x ∈ [0,1] and y ∈ [0,7] }

R = { (x,y): 0 ≤ x ≤ 1 and 0 ≤ x ≤ 7 }

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0}\int^{1}_{0} 4xye^{x^2 \ y} \ dx dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \begin {bmatrix} ye^{yx^2} \dfrac{4}{2y} \end {bmatrix}^1 _ 0 \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \begin {bmatrix} ye^{y1^2} \dfrac{4}{2y} - ye^{y0^2} \dfrac{4}{2y} \end {bmatrix}\ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \dfrac{4}{2}(e^y -1) \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \dfrac{4}{2}[e^y -1]^7_0 \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-(e^0 -0)]$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-1]$

$\mathbf{\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 (e^7 -8)}$

3 0

4 years ago

(Help!!!!!PLEASE!!!!!) The circumference of circle A is four times the circumference of circle B. Which statement about the area

lyudmila [28]

Answer:

B

Step-by-step explanation:

3 0

3 years ago

Which of the following is the equation of a circle with a radius of 10 cm and center at (–3, 6)?

vlabodo [156]

The equation of a circle is (x - h)² + (y - k)² = radius² where (h, k) is the ordered pair for the center.

(x - (- 3))² + (y - 6)² = 10²

(x + 3)² + (y - 6) = 100

5 0

4 years ago