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Nataly_w [17]
3 years ago
7

Ernesto Montenegro purchase a twelve 1000 Bonds having a quoted price of 100.75 he have to pay a 6% brokerage fee of the same pr

ice. What was the total cost of the bonds to the nearest cent
Mathematics
1 answer:
LenKa [72]3 years ago
5 0

Answer: $106.8

Step-by-step explanation:

Given that the bond price is $100.75

With a 6% brokerage fee

Find the total cost to the nearest cent

First, we calculate the 6% brokerage fee= 6/100 × 100.75

= 0.06 x 100.75

= 6.045

Total cost therefore is the sum of brokerage fee and the cost of bond

= $6.045 + $100.75

= $106.795

To the nearest cent = $106.8

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Solve the proportion x/6 = 4/5
luda_lava [24]

Answer:

x/6 = 4/5

x(5) = 6(4)

5x=24

x=24/5

Let me know if this helps!

3 0
2 years ago
Which inequality is represented by the graph?
PSYCHO15rus [73]

Answer:

<h2>NONE OF THEM</h2>

Step-by-step explanation:

From the graph,

y and x intercepts are (0,-3) and approximately (1.9,0) respectively

for A, C and D : y=53x-3

x intercept the value of y=0, and y intercept x=0.

x and y intercepts are (\frac{3}{53},0 ) and approximately (0,-3) respectively

for B: y=35x-5

 x and y intercepts are (\frac{3}{35},0 ) and approximately (0,-3) respectively

Therefore x and y intercepts for the lines y\leq 53x-3,y>53x-3, y and y they are not equal to the x and y intercepts of the given graph

3 0
2 years ago
An ant weigh 0.000003 kg. How much would 2 ants weigh, expressed in scientific notation?
Alik [6]
The correct answer is 6*10^-6 ..........................
5 0
3 years ago
Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are inches long, inches wide, and 8 inche
allochka39001 [22]

Answer:

- The number of small boxes that will fill the large box 1 = 64

- The number of small boxes that will fill the large box 2 = 56

Step-by-step explanation:

Complete Question

Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are 4.5 inches long, 4.5 inches wide, and 8 inches tall. To transport the items to an event, you want to know how many of the small boxes will fit in larger boxes. The larger boxes are available in two sizes. Large Box 1 is 24.25 inches long, 18 inches wide, and 24 inches tall. Large Box 2 is 20.5 inches long, 18.5 inches wide, and 24 inches tall. Both the small and large boxes must remain upright.

Solution

To know how many of the small boxes will fit in larger boxes, we need to obtain the volumes of the small box, large box 1 and large box 2.

Volume of a cuboid = L × W × H

For the small box,

Length = L = 4.5 inches

Width = W = 4.5 inches

Height = H = 8 inches

Volume of the small box = 4.5 × 4.5 × 8 = 162 in³

For large box 1,

Length = L = 24.25 inches

Width = W = 18 inches

Height = H = 24 inches

Volume of the large box 1 = 24.25 × 18 × 24 = 10,476 in³

For large box 2

Length = L = 20.5 inches

Width = W = 18.5 inches

Height = H = 24 inches

Volume of the large box 2 = 20.5 × 18.5 × 24 = 9,102 in³

The number of small boxes that'll fill the large box 1 = (10,476/162) = 64.667 = 64 small boxes (rounded down because the fraction cannot be forced into the large box 1.

The number of small boxes that will fill the large box 2 = (9,102/162) = 56.185 = 56 small boxes.

Hope this Helps!!!

7 0
2 years ago
Someone help me out with this need a quick respond
ser-zykov [4K]
To get the perimeter of a circle you need to use the formula 2xpix r
so 2x 3.1416 x 2= 12.56
so the third one counting from left to right
8 0
3 years ago
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