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3 years ago

15

Vladimir has 64 coins of .25 cents and 56 coins for .50 cents. What is the greatest number of identical packs of coins with no c

oins left over? Be sure to show you work

1 answer:

Zina [86]3 years ago

3 0

Answer:

56 coins for 0.50 cents

Step-by-step explanation:

You might be interested in

Use the chart to estimate the population density of Romania in 1997. Round to the nearest person.

STALIN [3.7K]

The population density of Romania in 1997 given its area and the population is 95 people per square km.

<h3>What is the population density?</h3>

Population density is the number of people that live in a square kilometer of an area. It is determined by dividing total population by total area.

Population density = population / area

22.600,000 / 237301 = 95 people per square km

To learn more about population density, please check:

brainly.com/question/3704320

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5 0

2 years ago

If you are solving for a system of equations algebraically and you end up with the equation 3=3... what is the solution to the s

Vlad1618 [11]

Answer:

All real numbers

Step-by-step explanation:

For example,

2x+3=2x+3

3=3

Or

2x=2x

x=x

It’s the same thing.

3 0

3 years ago

Read 2 more answers

1.)Solve the inequality. Graph the solution.

Anika [276]

Answer:

Part 1) $r\geq \sqrt[3]{-6}$

Part 2) $s>4/43$

Part 3) $z\leq6$

Part 4) $t>-7$

Part 5) $q$

Part 6) $p\geq -121/13$

Step-by-step explanation:

Part 1) $-r^{3}\leq 6$

Multiply by -1 both sides

$r^{3}\geq -6$

$r\geq \sqrt[3]{-6}$

$r\geq -1.817$

The solution is the interval -------> [-1.817,∞)

The solution in the attached figure N 1

Part 2) $-4>-43s$

rewrite

$-43s$

Multiply by -1 both sides

$43s>4$

$s>4/43$

$s>0.09$

The solution is the interval--------> (0.09,∞)

The solution in the attached figure N 2

Part 3) $z-2-6\leq-2$

combine like terms

$z-8\leq-2$

Adds 8 both sides

$z\leq-2+8$

$z\leq6$

The solution is the interval--------> (-∞,6]

The solution in the attached figure N 3

Part 4) $-2t-5$

Adds 5 both sides

$-2t$

$-2t$

Multiply by -1 both sides

$2t>-14$

Divide by 2 both sides

$t>-7$

The solution is the interval--------> (-7,∞)

The solution in the attached figure N 4

Part 5) $7(q+2)$

applying distributive property left side

$7q+14$

Subtract 14 both sides

$7q$

$7q$

Divide by 7 both sides

$q$

$q$

The solution is the interval--------> (-∞,-13)

The solution in the attached figure N 5

Part 6) $-13(p+9)\leq4$

applying distributive property left side

$-13p-117\leq4$

Adds 117 both sides

$-13p\leq4+117$

$-13p\leq 121$

Divide by -1 both sides

$13p\geq -121$

Divide by 13 both sides

$p\geq -121/13$

$p\geq -9.31$

The solution is the interval--------> [-9.31,∞)

The solution in the attached figure N 5

4 0

3 years ago

A line intersects the points (-5,1) and (-2,7). m=2 Write an equation in point slope form using the point (-5,1) y-[?] = ___(x-_

zalisa [80]

$(\stackrel{x_1}{-5}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{7}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{7}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{-2}-\underset{x_1}{(-5)}}} \implies \cfrac{7 -1}{-2 +5}\implies \cfrac{6}{3}\implies 2$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{2}(x-\stackrel{x_1}{(-5)})\implies y-1=2(x+5)$

7 0

2 years ago

You have five $1 bills, four $5 bills, six $10 bills, and three $20 bills you select a bill at random, without replacing the bil

Lostsunrise [7]

Answer:

Option B.

Step-by-step explanation:

It is given that,

Number of $1 bills = 5

Number of $5 bills = 4

Number of $10 bills = 6

Number of $20 bills = 3

So,

Total number of bills = 5 + 4 + 6 + 3 = 18

We need to find the probability of a 1 then a ten (without replacement).

Probability of selecting $1 bill in first draw $=\dfrac{5}{18}$

After one draw, the number of remaining bills is 18 - 1 = 17.

Probability of selecting $10 bill in second draw $=\dfrac{6}{17}$

The probability of a 1 then a ten is

$P=\dfrac{5}{18}\times \dfrac{6}{17}=\dfrac{5}{51}$

Therefore, the correct option is B.

6 0

3 years ago