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2 years ago

10

Nia plans to put enough money in the bank each week for 100 weeks so that the amount in the bank is always proportional to the w

eek number. On week 8, she had $20.00. On week 12, she had $30.00. How much money will be in the savings account on week 100?
$40
$66
$225
$250

2 answers:

dimaraw [331]2 years ago

7 0

250 in savings account

kobusy [5.1K]2 years ago

6 0

Nia will have $250 in the savings account

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Question 6: An experiment consists of throwing two six-sided dice and observing the number of spots on the upper faces. Determin

yaroslaw [1]

Answer:

(a) 0.25

(b) 0.944

(c) 0.444

(d) 0.167

Step-by-step explanation:

There are six possible outcomes for each die, which means that the number of possible outcomes is:

$n=6*6 = 36$

(a) In order for each die to show four or more spots they will both have to land on a four, five or six. The probability of this happening is:

$P(A) = \frac{3*3}{36}=0.25$

(b) There are only two possible outcomes for which the sum is three (1 and 2, or 2 and 1). The probability of the sum NOT being three is:

$P(B) = 1-\frac{2}{36}=0.944$

(c) If neither a one or a six must appear, then there are 4 possible outcomes for each die, the probability is:

$P(C) = \frac{4*4}{36}=0.444$

(d) For each one of the six possible numbers on the first die, there is only one on the second die for which the sum of the spots is 7, totaling six possible ways to sum 7:

$P(D) = \frac{6}{36}=0.167$

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2 years ago

A rat is trapped in a maze. Initially he has to choose one of two directions. If he goes to the right, then he will wander aroun

Brut [27]

Answer:

The expected number of minutes the rat will be trapped in the maze is 21 minutes.

Step-by-step explanation:

The rat has two directions to leave the maze.

The probability of selecting any of the two directions is, $\frac{1}{2}$ .

If the rat selects the right direction, the rat will return to the starting point after 3 minutes.

If the rat selects the left direction then the rat will leave the maze with probability $\frac{1}{3}$ after 2 minutes. And with probability $\frac{2}{3}$ the rat will return to the starting point after 5 minutes of wandering.

Let <em>X</em> = number of minutes the rat will be trapped in the maze.

Compute the expected value of <em>X</em> as follows:

$E(X)=[(3+E(X)\times\frac{1}{2} ]+[2\times\frac{1}{6} ]+[(5+E(X)\times\frac{2}{6} ]\\E(X)=\frac{3}{2} +\frac{E(X)}{2}+\frac{1}{3}+\frac{5}{3} +\frac{E(X)}{3} \\E(X)-\frac{E(X)}{2}-\frac{E(X)}{3}=\frac{3}{2} +\frac{1}{3}+\frac{5}{3} \\\frac{6E(X)-3E(X)-2E(X)}{6}=\frac{9+2+10}{6}\\\frac{E(X)}{6}=\frac{21}{6}\\E(X)=21$

Thus, the expected number of minutes the rat will be trapped in the maze is 21 minutes.

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2 years ago

Ralph Chase plans to sell a piece of property for $165000. He wants the money to be paid off in two ways - a short-term note at

KonstantinChe [14]

Answer:

The amount of the 10% note is $90,000 and the amount of the 7% note is $75,000

Step-by-step explanation:

Let the amount paid on the 7% note be $x and the amount paid on the 10% note be $y

we know that the sum of both is 165,000

Thus, mathematically

x + y = 165,000 •••••••(i)

Now let’s take a look at interest paid

For the 7%;

7/100 * x = 0.07 * x = 0.07x

For the 10%

10/100 * y = 0.1 * y = 0.1y

The addition of both gives $14,250

0.07x + 0.1y = 14,250 •••••(ii)

We now solve both equations simultaneously

From equation 1, we can have;

x = 165,000 -y

we can substitute this into ii

0.07(165,000-y) + 0.1y = 14,250

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0.03y = 14,250-11,550

0.03y = 2700

y = 2700/0.03

y = $90,000

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x = 165,000-y

x = 165,000 -90,000

x = $75,000

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Answer:

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If Mark sold 2 7/8 gallons and Regan sold 2/3 as much, then Regan sold

$2\frac{7}{8}*\frac{2}{3}$

I teach my algebra students to change that mixed fraction into an improper to make the multiplication easier:

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0 0

2 years ago