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1 year ago

Is the U.S. ever on sale to canadians?

Mathematics

1 answer:

Goshia [24]1 year ago

7 0

Answer:

No because Us is powerful country canada is also powerful but U S ever sale to candians

You might be interested in

Find a third-degree polynomial equation with rational coefficients that has roots -4 and 6 + i.

dolphi86 [110]

Answer:

x³ - 8x² - 11x + 148

Step-by-step explanation:

Given that x = 6 + i is a root then x = 6 - i is also a root

Complex roots occur as conjugate pairs.

The factors are therefore (x - (6 + i)) and(x - (6 - i))

Given x = - 4 is a root then (x + 4) is a factor

The polynomial is the product of the factors, that is

p(x) = (x + 4)(x - (6 + i))(x - (6 - i))

= (x + 4)(x - 6 - i)(x - 6 + i)

= (x + 4)((x - 6)² - i²)

= (x + 4)(x² - 12x + 36 + 1)

= (x + 4)(x² - 12x + 37) ← distribute

= x³ + 4x² - 12x² - 48x + 37x + 148

= x³ - 8x² - 11x + 148

3 0

3 years ago

I need help on 14 please

Fynjy0 [20]

2+x=50

That is what I would do. I hope this helps!

8 0

3 years ago

Read 2 more answers

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago

Solve for xxx. Your answer must be simplified. \dfrac x{-6}\geq-20 −6 x ≥−20

Tems11 [23]

Answer:

x ≤ - 27

Step-by-step explanation:

$\frac{x}{-9}\ge \:3\\\\\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}\\\\\frac{x\left(-1\right)}{-9}\le \:3\left(-1\right)\\\\Simplify\\\\\frac{x}{9}\le \:-3\\\\\mathrm{Multiply\:both\:sides\:by\:}9\\\\\frac{9x}{9}\le \:9\left(-3\right)\\\\\mathrm{Simplify}\\\\x\le \:-27$

7 0

3 years ago

Find the average rate of change (-4,-7)(-2,3)

Basile [38]

The average rate of change is m=5.
Hope this helped.

6 0

2 years ago