What is the value of 3√4096? A. 512 B. 64 C. 16 D. 8

Graph f(x) = x and g(x) = -1/3x+2 on the same coordinate plane.

Vesnalui [34]

It’s not too hard,

I hope that helps.

6 0

3 years ago

2x^2=9-3x Find the factors

d1i1m1o1n [39]

Answer:

x = -3, $\frac{3}{2}$

Step-by-step explanation:

The given quadratic equation is: $2x^2 = 9 - 3x$

This can be written as: $2x^2 + 3x - 9 = 0$

To solve a quadratic equation of the form $ax^2 + bx + c = 0$ we use the formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, a = 2; b = 3; c = - 9

Therefore, the roots of the equation are:

$x = \frac{- 3 \pm \sqrt{9 - 4(2)(-9)}}{2(2)}$

$\implies x = \frac{-3 \pm \sqrt{81}}{4}$

$\implies x = \frac{-3 \pm 9}{4}$

We get two values of 'x', viz.,

x = $\frac{-3 + 9}{4}$ and $\frac{- 3 - 9}{4}$

$\implies x = \frac{6}{4} \hspace{5mm} \& \hspace{5mm} \frac{-12}{4}$

⇒ x = -3, 3/2

Since the factors of the quadratic equation is asked, we write it as:

(x + 3)(x - $\frac{3}{2}$ ) = 0

because, if (x - a)(x - b) are the factors of a quadratic equation, then 'a' and 'b' are its roots.

Multiply (x + 3) and (x - $\frac{3}{2}$ to see that this indeed is the given quadratic equation.

5 0

3 years ago

Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. Upper H 0: pequals0.6 versus U

kow [346]

Answer:

Step-by-step explanation:

From the given information:

The null and the alternative hypothesis can be well written as:

$H_o:P=0.6$

$H_1:P>0.6$

Given that:

n = 200

x = 135

Alpha ∝ = 0.05 level of significance

Then;

⇒ $n \times p\times (1-P)$

= 200 × 0.6 × (1 -0.6)

= 200 × 0.6 × 0.4

= 48 ≥ 10

The sample proportion $\hat P = \dfrac{x}{n}$

$= \dfrac{135}{200}$

= 0.675

The test statistics $Z = \dfrac{\hat P - P}{\sqrt{ \dfrac{P(1-P)}{n} }}$

$Z = \dfrac{0.675 - 0.6}{\sqrt{ \dfrac{0.6 \times 0.4}{200} }}$

$Z = \dfrac{0.075}{\sqrt{ \dfrac{0.24}{200} }}$

Z = 2.165

The P-value = P(Z > 2.165)

= 1 - P(Z < 2.165)

From the z tables

= 1 - 0.9848

= 0.0152

Reject the null hypothesis since P-Value is lesser than alpha. ( i.e. 0.0152 < 0.05).

Thus, there is enough evidence to conclude that the value of the population proportion is greater than 0.6

7 0

3 years ago

William has been contracted to paint a school classroom. The classroom is 20 m long, 15 m wide and 5 m high. There are four wind

valina [46]

Answer:

Step-by-step explanation:

l -> length

b -> width

h -> height

Find the area of four walls and ceiling. then subtract the area of four windows and a door form that area.

Area of four walls + ceiling = 2( lh + bh) +lb

= 2*(20*5 + 15*5) + 20*15

= 2( 100 + 75) + 300

= 2* 175 + 300

= 350 +300

= 650 sq m

Area of window = 2 *3 = 6 sq.m

Area of four windows = 4*6 = 24 sq.m

Area of door = 2 * 1 = 2 sq.m

Area of four walls excluding 4 windows and door = 650 - 24 - 2 = 624 sq.m

Cost of painting = 624 * 6.50

= $ 4056

3 0

3 years ago

Read 2 more answers

Can someone actually help me with these problems? They are due tomorrow.

Ainat [17]

Answer:

Hope this helps you

Step-by-step explanation:

43. A "Square" cornfield, which means all 4 sides has the same lenght BxH

164.000 ft^2=404.96 NO

156.816 ft^2=396 Yes

174724 ft^2=418 Yes

215908 ft^2=464.66 NO

44. it's the same here, if the area is 81. then each side of the square is 9

1st figure: it has 12 sides x9 =108

2nd figure: it has 10 sides x9=90

3rd figure: has also 12 sides x9=108

45. 13^3= 13x13x13 = 2197

46. 25^2= 25x25 = 625

47. 15^3= 15x15x15 = 3375

48. 34^2= 34x34 = 1156

49. 5 root121= 5 (11)=55

50. -6.root36= - 6 (6)= -36

51. 10 root3 (8) = 10(2) = 20

52. -4. root 144= -4(12) = -48

3 0

2 years ago