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Tcecarenko [31]

1 year ago

15

Eugene and three of his friends went out to eat. They decided to split the bill evenly. Each person paid $11. What was the total

bill?

1 answer:

Elena-2011 [213]1 year ago

8 0

Answer:

Step-by-step explanation:

Je calcule la facture totale

11x4=44

La facture totale d'Eugène et ses trois amis était de 11$

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A bag contains 6 yellow marbles and 18 red. If a representative sample contains 2 yellow marbles, then how many red would you ex

kramer

Answer:

6 red marbles

Step-by-step explanation:

We must first find our multiplier since a sample represents a population (the whole thing).

6/2 = 3, so 3 * 2 = 6.

I will use the value of 6/2, aka 3 and use that as my divisor.

18/3 = 6.

Therefore, you would expect to have 6 red marbles.

3 0

3 years ago

a number is selected at random from set (2,3,45,6,7,8,9,and 10). which event covers the entire sample space of this experiment ?

ki77a [65]

I'm not sure I understand the question. Is this worded correctly?

8 0

3 years ago

Your job is to sort and sack marbles for sale. A bag contains 44 marbles, some red and some green. If there are 11 red marbles,

julia-pushkina [17]

Answer:

Option C is correct

The ratio of green to red marbles is, 3 : 1

Step-by-step explanation:

As per the statement:

A bag contains 44 marbles, some red and some green.

⇒Total marbles = 44

If there are 11 red marbles.

then;

Green marbles = Total marbles - red marbles.

Substitute the given values we have;

Green marbles = 44 - 11 = 33

We have to find the ratio of green to red marbles.

$\frac{\text{Green marbles}}{\text{Red marbles}} = \frac{33}{11} = \frac{3}{1} = 3 : 1$

Therefore, the ratio of green to red marbles is, 3 : 1

8 0

3 years ago

the perimeter of a picture frame is 24 inches if the width of the picture frame 5 inches what is the length of the picture frame

Ket [755]

You would divide 24 by 5 so the length would end up being: 4.8 in
Hope it helps!

7 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago