Which expression is equivalent to: (5⁹ × 5⁻⁷)⁻² ?

AleksAgata [21]

12 times 3.14 = 37.68 CM
C=D x 3.14

5 0

3 years ago

Three friends Kofi,Yaw and Musa are in partnership business in which Kofi and Yaw contributed 5/9 and 2/15 of the cost of 14000.

wlad13 [49]

Answer:

(a) $Total = 45000$

(b) Yaw, Musa and Kofi

Step-by-step explanation:

Given

$Kofi =5/9$

$Yaw = 2/15$

$Musa = 14000$

Solving (a): The cost of the business

First, we solve for the fraction of Musa

$Musa + Kofi + Yaw = 1$

$Musa = 1 - (Kofi + Yaw )$

$Musa = 1 - (5/9 + 2/15)$

$Musa = 1 - (\frac{25+6}{45})$

$Musa = 1 - \frac{31}{45}$

$Musa = \frac{45 - 31}{45}$

$Musa = \frac{14}{45}$

So, we have:

$\frac{14}{45} * Total = 14000$

Make Total the subject

$Total = 14000 * \frac{45}{14}$

$Total = 1000 * 45$

$Total = 45000$

Solving (b): Partners in ascending order

First, represent the fractions as decimals

$Kofi =5/9$ $= 0.5556$

$Yaw = 2/15$ $= 0.1333$

$Musa = \frac{14}{45}$ $= 0.3111$

From the conversion above, the least is 0.1333 (Yaw), then 0.3111 (Musa), then 0.5556 (Kofi).

<em>So, the order is: Yaw, Musa and Kofi</em>

6 0

3 years ago

Rewrite the following equation in standard form.<br><br> y= -5/8x− 7

Colt1911 [192]

Answer:

Step-by-step explanation:

y= -5/8x-7

To write in standard form get all the terms to one side

y+5/8x+7=0

I hope this is correct

Pls don't be mad if it is wrong

6 0

3 years ago

An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bu

notsponge [240]

Answer:

1. 90% 2. 10% 3. 50%

Step-by-step explanation:

Standard Deviation (σ) = 50 days

Average/Mean (μ) = 300days

Probability that it would last more than 300 days = P(Bulb>300 days)

We will assume there are 365 days in a year.

P(Bulb>300 days) implies that the bulb would

Using the normal equation;

z = standard/normal score = (x-μ)/σ where x is the value to be standardized

P(Bulb>300 days) implies x = 365 days

Therefore z = (365-300)/50 = 1.3

Using the normal graph for z=1.3, probability = 90%

2. P(Bulb<300 days) = 1 - P(Bulb>300 days)\

P(Bulb<300 days) = 1 - 0.9

P(Bulb<300 days) = 10%

3. P(Bulb=300 days) implies z=0 since x=300

Using the normal graph for z=0, probability =50%

6 0

3 years ago

Can someone please help me with this geometry homework?

Leno4ka [110]

The distances between the two points, in units, are given as follows:

7. $\sqrt{2}$

8. $2\sqrt{17}$

9. $3\sqrt{2}$

10. $3\sqrt{5}$

The midpoints are given as follows:

11. (7,-4.5).

12. (2,4).

13. (7,0.5).

14. (5.5,-1.5).

<h3>What is the distance between two points?</h3>

Suppose that we have two points, $(x_1,y_1)$ and $(x_2,y_2)$ . The distance between them is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For item 7, the points are (-3,-2) and (-4,-3), hence the distance is:

$D = \sqrt{(-4 + 3)^2 + (-3 + 2)^2} = \sqrt{2}$

For item 8, the points are (-3,1) and (5,3), hence the distance is:

$D = \sqrt{(5 + 3)^2 + (3 - 1)^2} = \sqrt{68} = 2\sqrt{17}$

For item 9, the points are (-3,0) and (0,-3), hence the distance is:

$D = \sqrt{(0 + 3)^2 + (-3 - 0)^2} = \sqrt{18} = 3\sqrt{2}$

For item 10, the points are (1,-2) and (4,4), hence the distance is:

$D = \sqrt{(4 - 1)^2 + (4 + 2)^2} = \sqrt{45} = 3\sqrt{5}$

<h3>What is the midpoint concept?</h3>

The midpoint between two points is the halfway point between them, and is found using the mean of the coordinates.

Hence, for item 11, the midpoint is given by:

[0.5(9 + 5), 0.5(1-10)] = (7,-4.5).

For item 12, the midpoint is given by:

[0.5(-5 + 9), 0.5(7 + 1)] = (2,4).

For item 13, the midpoint is given by:

[0.5(9 + 5), 0.5(9 - 8)] = (7,0.5).

For item 14, the midpoint is given by:

[0.5(10 + 1), 0.5(4 - 7)] = (5.5,-1.5).

More can be learned about the distance between two points at brainly.com/question/18345417

#SPJ1

8 0

1 year ago