Reynard and his friends visited an orchard on Sunday. They picked 12.34 pounds of apples and

What is the first quartile and third quartile of this data set: 34, 34, 36, 39, 50, 41, 45, 47, 40?

o-na [289]

Answer:

whatever

Step-by-step explanation:

the first quartile is between 34 and 36 = 35

the third quartile is between 45 and 47 = 46

5 0

2 years ago

In trei rezervoare sunt 1672 l de benzina dacã in primele doua rezervoare sunt 123100 cl iar in ultimele doua sunt 15 hl sa se a

sesenic [268]

the answer is

615500

hope this helps

6 0

3 years ago

Two equations are given below:

stealth61 [152]

a is given to us so just plug a into the first equation:

b-3 - 3b =9

Add 3 to both sides:

b-3b=12

Combine like terms:

-2b=12

Divide by -2 to get b by itself:

b=-6

The only answer with b as -6 is the first one, (-9,-6)

8 0

2 years ago

Read 2 more answers

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago

The ratio of the perimeters of two similar triangles is 4:3. What are the areas of these triangles if the sum of their areas is

emmasim [6.3K]

Answer:

the areas of these triangles are 83.2cm² and 46.8cm²

Step-by-step explanation:

1. If the triangles are similar and the ratio of the perimeter ois 4:3, then the areas are in the following ratio:

4²:3²

16:9

2. The sum of their areas is 65 cm², then, you can calculate the area of the larger triangle as following:

130(16/16+9)

130(0.64)

=83.2cm²

3. The area of the smaller triangle is:

130(9/16+9)

130(0.36)

46.8cm²

<u>Hope this help</u>s

6 0

2 years ago