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3 years ago

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7. A business begins with $1,000,000 from its owners. It places $100,000 into a bank account which it wishes to maintain as a ca

sh reserve. It needs to purchase $1,000,000 of equipment and materials. How much does it need to borrow

1 answer:

joja [24]3 years ago

8 0

Answer:

The amount need to borrow a owner is $100,000

Step-by-step explanation:

The total amount a owner having to start a business = $1,000,000.

He wishes to place $ 100,000 into bank account to maintain as cash reserve.

Now the amount he left with to spent in business purposes = $1,000,000 - $100,000 = $ 900,000.

He needs to purchase equipment and material of $1,000,000

But he is left with only $900,000.

So the amount he has to borrow to fulfill his business requirement is

= $1,000,000 - $900,000 = $100,000

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MathWizard likes to play a fun number trick on her friends. She tells them to think of a number. She then tells them to subtract

pantera1 [17]

Follow the steps for 8
(1) 8 * 2 = 16
(2) 16 * 5 = 80
Note: you could just combine the first 2 steps and multiply by 10 to start with. Much easier: just add a 0 onto 8.
(3a) 12 * 8 = 96
(3b) 96 - 24 = 72
(3c) So you want to take 1/2 72 which is 36
(4) 36 + 80 = 116

Let's try this with x.
(2) x * 10 = 10x
(3) 1/2 (12*x - 24)
(3) 10x + 1/2 (12x - 24)
(3) 10x + 6x - 12
(4) 16x - 12

Problem B
Math wiz would add 12 to the number given to her and then divide by 16. That's pretty complicated to do quickly.

For example Let the original number be 1221
(2) Multiply by 10 12210
(3) Take 1/2 the difference between 12 times 1221 and 24 which is 1/2(14652 - 24) = 7314
And add this result to 12210
7314 + 12210 = 19524

Tell her the number. She has to add 12
19524 + 12 = 19536 And divide by 16
to get 1221.
Pretty complex even for a wizard.

6 0

3 years ago

If i have zero friends and get zero more friends how many friends do i have 100 points

horrorfan [7]

Answer:

You have 0 friends just like me :D

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3 years ago

Read 2 more answers

Help (this for a friend)<br> a 62.1<br> b 117.4<br> c 122.1<br> d 164.5

kramer

Answer:

c. 122.1 cm

Step-by-step explanation:

so l * w + $\frac{\pi r^2}{2}$

since 6 cm is the diameter

6/2 = 3

the radius (or r) is 3

so 18 * 6 + $\frac{3.14 * 3*3}{y}$

108 + $\frac{3.14*9}{y}$

108 + $\frac{28.26}{2}$

108 + 14.13 = 122.13

simplified: 122.1 cm

hope this helps:)

4 0

2 years ago

Read 2 more answers

John Tucker is interested in purchasing stock. He wants to determine the costs.

Delvig [45]

Answer: a0 =$97.5 ; a1 =$87.96 ; a2=$282.48; a3=$1,722.5 ; a4=$1,553.96 ; a5=$4,990.48

Step-by-step explanation:

Hi, to solve this problem we have to multiply each total by the commission in decimal form (6 / 100= 0.06).

So:

$1625 x 0.06 = $97.5

$1,466 x 0.06 =$87.96

$4,708 x 0.06 =$282.48

And finally, to obtain the total costs we have to add to each total the commission of the total.

$1625 + $97.5 = $1,722.5

$1,466 +$87.96 = $1,553.96

$4,708 +$282.48 = $4,990.48

Feel free to ask for more if it´s necessary or if you did not understand something.

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3 years ago

Read 2 more answers

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(c) 0.0039

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0

3 years ago