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3 years ago

Anyone here a kpop stan?

Mathematics

2 answers:

Fiesta28 [93]3 years ago

5 0

Oop umm me... haha why?

klasskru [66]3 years ago

3 0

Nooooo but why ? Is it important

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Which numbers should go in the shaded overlap? <br> Factors of 30<br> Factors of 20

babymother [125]

Both of these numbers have 2 and 5 as their factors

5 0

3 years ago

What expression is equal to 75. Squared?

Murljashka [212]

The answer to 75 squared is 5,625.

8 0

4 years ago

Find the area of the shaded sector of the circle. Leave in terms of pi.

qaws [65]

Answer:

$a=10.88\pi$

Step-by-step explanation:

using the formula below, the central angle is 20° (bc 360-180-160=20)

radius is 14 (28/2=14)

$\frac{20}{360} =\frac{a}{\pi r^2} \\ \frac{1}{18} =\frac{a}{\pi 14^2}$

$\frac{1}{18} =\frac{a}{196\pi}$

$a=\frac{1*196\pi}{18} \\a=1*10.88\pi\\a=10.88\pi$

3 0

3 years ago

Anna, Beth and Colleen are softball players. This season, Anne had 65 base hits in 160 at-bats. Beth batting average was 0.399,

zheka24 [161]

Answer:

Anne is the best hitter as she has the best hitting average of 0.406

Step-by-step explanation:

Anne hits 65 bases out of 160 at - bats.

Beth's betting average = 0.399

Collin's hit = 40%

Now, we need to convert all the three hitting rates into same parameters to compare them with each other.

Average hit rate of Anne = $\frac{\textrm{Total numberof base}}{\textrm{Total number of at - bats}}$

$= \frac{65}{160} = 0.406$

⇒The betting average of Anne = 0.406

Now, The hit - rate of Collin = 40%

Converting the percentage rate into decimal, we get the hit rate of Collin

= $\frac{40}{100} = 0.40$

Hence, we get that Hitting Average of :

<u>Anne = 0.406</u>, Beth = 0.3999 and Collin = 0.400

Hence, from the above data Anne is the best hitter as she has the best hitting average of 0.406.

6 0

3 years ago

Read 2 more answers

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago