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pav-90 [236]
2 years ago
6

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
LekaFEV [45]2 years ago
8 0
It’s A!!! Lol dneneenne
bezimeni [28]2 years ago
7 0
Pick a pick a dont actually
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A total of 15000 is invested at an annual percentage rate of 12.5%. Find the accumulated amount after 3 years if it is compounde
aleksklad [387]

Answer:

24542.26 is the final balance

7 0
3 years ago
A homemade loaf of bread turns out to be a perfect cube. Five slices of bread, each 0.6 inch thick, are cut from one end of the
erastovalidia [21]

A cube has all sides of equal length.

So let the side be x

that will be length = width = height = x  

Now, when we cut the loaf into 5 equal slices, the length and height of the slices are the same as the original loaf.  But the width will become less with each slicing. Here the width corresponds to the thickness of the slices.

For each slice the dimensions are -

Length = x

height = x

width = 0.6

Volume of the original loaf is x³.

Volume of  5 slices is 5(x * x * 0.6).  

Volume that is left after the cutting the slices is 235 inch³.

So, equation of the volume is=

x³ - 5(0.6x²) = 235

Solve for x using this equation.

x³ - 3x² = 235

x³ - 3x² - 235 = 0

Using a graph calculator to find the roots, we get

x = 7.35

Hence the dimensions of the original loaf is 7.35 inches for each side.

Please find attached the graph.

7 0
3 years ago
g A trial jury of 6 people is selected from 20 people: 8 women and 12 men. What is the probability that the jury will have an od
cricket20 [7]

Answer: P(odd) = 0.499

Step-by-step explanation:

Given:

Total number of people = 20

Number of men = 12

Number of women = 8

Number of jury to be selected = 6

For the jury to have an odd number of women. it must have either of the three.

1. 1 woman , 5 men

2. 3 women, 3 men

3. 5 women, 1 man

The total possible ways of selecting the 6 people jury is;

N = 20C6 = 20!/6!(20-6)!

N = 38760

The possible ways of selecting;

Case 1 : 1 woman, 5 men

N1 = 8C1 × 12C5

N1 = 8 × 792 = 6336

Case 2 : 3 women , 3 men

N2 = 8C3 × 12C3

N2 = 12320

Case 3 : 5 women, 1 man

N3 = 8C5 × 12C1

N3 = 672

P(Odd) = (N1+N2+N3)/N

P(odd) = (6336+12320+672)/38760

P(odd) = 19328/38760

P(odd) = 0.499

6 0
3 years ago
Find two consecutive integers whose sum is 73. Which of the following equations could be used to solve the problem?
max2010maxim [7]
2x  + 1 = 73

2x =  72
x = 36  


the 2 integers are 36 and 37

Its  b
8 0
3 years ago
Read 2 more answers
Someone give me the answer to this problem please ☺️☺️
Softa [21]
SA = 2 * pi * r^2 + 2 * pi * r * h
r = 1
h = 4
pi = 3.14
now sub
SA = 2(3.14)(1^2) + 2(3.14)(1)(4)
SA = 6.28(1) + 6.28(4)
SA = 6.28 + 25.12
SA = 31.4 sq ft

4 0
3 years ago
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