All categories

3 years ago

There are 30 actors in the school play. 60% of the actors are girls. How many actors are boys?

Mathematics

2 answers:

photoshop1234 [79]3 years ago

8 0

Answer:

Step-by-step explanation:

If 60% are girls, 40% are boys

100 - 60 = 40

40% of 30

(40/100)×30

= 12

steposvetlana [31]3 years ago

7 0

Answer: 12

Step-by-step explanation:

If 60% of the actors are girls, then 60/100×30= 18 girls

Subtract 18 girls from the total number of actors

30-18= 12 boys

You might be interested in

One number is 5 more than twice another and their sum is 23.

icang [17]

y = 2x + 5

x + y = 23

3 0

3 years ago

Read 2 more answers

This is due like right now, please help me!!!!

Varvara68 [4.7K]

$\bold{\huge{\orange{\underline{ Solution }}}}$

$\bold{\underline{ Given \: Rules}}$

The sum of the number in each of the four rows is the same
The sum of the numbers in each of the three columns is the same
The sum of any row does not equal the sum of any column

$\bold{\underline{ Let's \: Begin}}$

According to the Second rule :-

$\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }$

$\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}$

According to the first rule :- 

$\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}$

$\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}$

From ( 1 ) we got :-

$\sf{ 158 + b = 166 + a, 156 + d = 166 + a }$

$\sf{ b = 166 - 158 + a, d = 166 - 156 + a }$

$\sf{ b = 8 + a, d = 10 + a ...(3)}$

Subsitute (3) in ( 2 ) :-

$\sf{229+a = 161+8+a+c = 168+10+a+e}$

$\sf{ 229+a = 169+a+c = 178+a+e}$

We can write it as :-

$\sf{ 229+a = 169+a+c \:or\:229+a = 178+a+e}$

$\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}$

$\sf{ c = 60 \: and \: e = 51 }$

Subsitute the value of c and e in ( 1 ):-

$\sf{ 158 + b = 156 + d = 166 + a = 78 + 60 + 51 }$

$\sf{ 158 + b = 156 + d = 166 + a = 189}$

Now, 

$\sf{ For \: b, 158 + b = 189 }$

$\sf{ b = 189 - 158 }$

$\sf{ b = 31}$

$\sf{ For \: d , 156 + b = 189 }$

$\sf{ d = 189 - 156 }$

$\sf{ d = 33}$

$\sf{ For \: a, 166 + a = 189 }$

$\sf{ a = 189 - 166 }$

$\sf{ a = 23 }$

Hence, The value of a, b, c, d and e is 23, 31 ,60 ,33 and 51 .

8 0

3 years ago

Two normal six-sided dice are thrown 400 times. How many double results are there likely to be?

Verizon [17]

Answer:

66.66

Step-by-step explanation:

7 0

3 years ago

Can someone help me with this problem I really don’t know how to work it out and I’m getting frustrated please help..

11Alexandr11 [23.1K]

Answer:

1 cake and 6 pies

Step-by-step explanation:

Let c represent the number of cakes Alice and George can make. Likewise, let p represent the number of pies.

Then the amount of flour they need (in cups) is ...

3c + 2p . . . . cups of flour needed

And the amount of sugar needed is ...

2c + 1.5p . . . cups of sugar needed

They want to use all 15 cups of flour they have, and all 11 cups of sugar, so the amounts above need to match the amounts on hand:

3c + 2p = 15

2c + 1.5p = 11

We can solve these equations various ways. One of my favorite is to use a graphing calculator. (See attached) It tells us that Alice and George can make 1 cake and 6 pies with the flour and sugar they have.

_____

If you want to solve these equations algebraically, there are several methods for that, too. Here, we can multiply the first one by 2/3 and subtract the result from the second one:

(2c +1.5p) -(2/3)(3c +2p) = (11) -(2/3)(15)

2c +3/2p -2c -4/3p = 11 -10 . . . . . eliminate parentheses

1/6p = 1 . . . . . . . . . . . . . . . . . . . . . .collect terms

p = 6 . . . . . . . . . . . . . . . . . . . . . . . . multiply by 6

Substituting into the first equation gives ...

3c +2·6 = 15 . . . substitute 6 for p

3c = 3 . . . . . . . . subtract 12

c = 1 . . . . . . . . . . divide by 3

These values tell us Alice and George can make 1 cake and 6 pies with the flour and sugar they have.

7 0

3 years ago

(Square root 5-8)(-1+3square root 5)

RUDIKE [14]

Answer:

Step-by-step explanation:

This is what I got in the calculator

4 0

3 years ago