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3 years ago

8 times 604 use subtraction

Mathematics

1 answer:

tiny-mole [99]3 years ago

4 0

Answer:

4,832

Step-by-step explanation:

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Last week, Bill ran 900 meters on each of 3 days. Between last week and this week, bill wants to run a total of 6 kilometers. Ho

laila [671]

For the 3 days he ran a total of : 3 * 900 = 2700 meters.

1000 meters = 1 kilometers

2700 meters = 2700 / 1000 = 2.7 kilometers.

Between last week and this week, Bill ran a total of 6 kilometers.

How far he ran this week = 6 - 2.7 = 3.3 kilometers.

Option C.

7 0

3 years ago

22. Sue can paint a chair in 3 hours. Jerry can paint the same size chair in 2 1/2

iVinArrow [24]

I think it’s 55 hours, about more than 2 days

6 0

3 years ago

Cardinals jerseys come in three different sizes, medium, large and extra large. A medium is x dollars, a large is $9 more than t

kicyunya [14]

Answer:

MEDIUM JERSEY =$21

Step-by-step explanation:

MEDIUM=$X ,LARGE =$X+9, EX LARGE=$X+15

9(X+9)+4(X+15)=414

9X+81+4X+60=414

13X=414-141

13X=273

X=21

8 0

3 years ago

The length of a field in yards is a function f(n) of the length n in feet. Write a function rule for th

lara [203]

Answer:

$f(n) = \frac{n}{3}$

Step-by-step explanation:

Recall that:

$3ft = 1 \: yd$

This implies that:

$6ft = 2yds$

$9ft = 3yds$

In general:

$n \: ft = \frac{n}{3} \: \: yds$

We can therefore write the function rule:

$f(n) = \frac{n}{3}$

3 0

3 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.

3 0

4 years ago