Ask question

Ask question

All categories

2 years ago

12

Attendance at the big game was only 80% of last years attendance, but that still means 12,000 people showed up. how many people

showed up last year?

1 answer:

astraxan [27]2 years ago

8 0

12000*100/80=15000 coz 12000 is 80% that's why you divide by 80 and multiply by 100

You might be interested in

In the figure below, BD and AC are diameters of circle P.

elena-14-01-66 [18.8K]

Answer:

$arc\ DC=30^o$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

The point P is the center of the circle (the drawing is not a scale )

step 1

Find the measure of angle ∠CPD

we know that

$m\angle CPD+150^o=180^o$ ---> by supplementary angles (form a linear pair)

$m\angle CPD=180^o-150^o=30^o$

step 2

Find the measure of arc DC

we know that

<u><em>Central angle</em></u> is the angle that has its vertex in the center of the circumference and the sides are radii of it

so

$arc\ DC=m\angle CPD$ ----> by central angle

therefore

$arc\ DC=30^o$

5 0

2 years ago

346772.99378 x 222222666886=

nordsb [41]

Answer:

7.706081948 lowecase10uppercase16

Step-by-step explanation:

I hope it helps u.

6 0

2 years ago

10d - 6d = 4 find what d equals

Nutka1998 [239]

Answer:

$\huge\boxed{\sf d = 1}$

Step-by-step explanation:

10d - 6d = 4

Take d common

d (10 - 6) = 4

d (4) = 4

4d = 4

Divide both sides by 4

d = 1

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3>

5 0

2 years ago

Read 2 more answers

Worth lots of points :) Find the area of the parallelogram below.

Lostsunrise [7]

23.8 inches squared if im wrong sorry but what i did was i added 7.4 plus 7.4 cause there are two of the same side then did the same with 4.5

7 0

2 years ago

If the temperature after 4.4 degree drop was -2.5 degrees Celsius what was the temperature at noon

Lostsunrise [7]

Answer:

1.9°C

Step-by-step explanation:

Let X be the temperature at noon and -2.5°C be the current temperature after the drop.

-We calculate temperature at noon as:

$\bigtriangleup =t_o-t_n\\4.4=X--2.5\\\\X=4.4-2.5\\\\=1.5\textdegree C$

Hence, the temperature at noon was 1.9°C

5 0

2 years ago