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3 years ago

If 28 parrots and 79 bunnies were in the same shelter what is the rate of parrots to bunnies?

Mathematics

2 answers:

nalin [4]3 years ago

6 0

Put both values into a ratio

P:B
28:79

ivanzaharov [21]3 years ago

5 0

Answer:

79/28

Step-by-step explanation:

i have no clue-

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Need help with easy/quick probability problems. Do 1-6 and explain how to do one of them

alexandr1967 [171]

Answer:

P(algebra, then statistics) = 7/80

Step-by-step explanation:

The total number of books is 7 + 8 + 5 = 20.

1. P(algebra, then statistics)

The probability that the first book is an algebra book is 7/20.

Carlos then replaces the book, so the total is still 20.

The probability that the second book is a statistics book is 5/20.

Therefore, the total probability is:

P(algebra, then statistics) = 7/20 × 5/20

P(algebra, then statistics) = 7/80

8 0

4 years ago

I need help with this problem<br><br><br><br>will mark as brainliest

Arada [10]

Answer:

The shop can cut the cost of production per sandwich to $3.75. The revenue function would then be defined as R(x)=9x.

Step-by-step explanation:

We want to make as big profit as possible by making as little sandwiches as possible. This means that we want the profit, which is $revenue- production\: cost$ to be as big as possible.

For the first choice the profit per sandwich is $8.25-4=4.25$ dollars.

For the second choice the profit per sandwich is $9-3.75=5.25$ dollars.

For the third choice the profit per sandwich is $8.5-3.5=5$ dollars.

We see here that for the second choice the profit is greatest, therefore this choice is most suitably correct.

7 0

3 years ago

The measures of two sides of a triangle are 15 and 16. between what two numbers must the measure of the third side fall?

QveST [7]

If it is a right triangle and you given me adjacent and opposite the formula is
${r}^{2} = {x}^{2} + {y}^{2}$
third side = 21.93

if its not right triangle you need the angle to solve

6 0

3 years ago

A dance school charges a registration fee in addition to a fee per lesson. The equation LaTeX: y=30x+25 y = 30 x + 25 represents

klasskru [66]

The equation is 30x + 25
since the x is with the 30, that would be the cost per lesson, because you would multiply 30 by the number of lessons(x)

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