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3 years ago

Help me ASAP this answer is a lil tough

Mathematics

2 answers:

marta [7]3 years ago

7 0

The right choice would be c

Gnom [1K]3 years ago

5 0

I’ll help you but you’re from my class and that’s cheating so no

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The ratio of boys to girls in Mrs. Ronilo's math class is 3 to 2. What percent of the class is boys? A) 33% B) 40% C) 60% D) 66%

nikdorinn [45]

Answer:

C) 60.

Step-by-step explanation:

Okay, I do math a little bit different than other kids... No worries though, its still correct.

First, find out how many times five can be put into a hundred. You should have gotten 20.

Now, we can use the twenty we got. Multiply 20 by the ratio of the boys. So 20 * 3 = 60.

Girls would be

20 * 2 = 40.

Those two numbers add up to a hundred so its all good. ^-^

Hope I helped!!

8 0

3 years ago

I got 10 min to submit pls I need it asap

babymother [125]

Answer:

We just need the equations to equal the same, which is y.

5+1.10x=y

3+1.50x=y

5+1.10x= 3+1.50x

Let's get x by itself by subtract 3 from both sides.

2+1.10x= 1.50x

Now subtract 1.10x from both sides.

2= 0.4x

Now divide each side by 0.4

5=X

So, at 5 games they're the same cost. Let's find the cost.

5+1.1(5)= 5+5.5=10.50

3+1.5(5)=3+7.5= 10.50

So, at 5 games, they are $10.50.

Hope this helps! :)

5 0

2 years ago

Complete the following sentence.

Fantom [35]

Answer:

it is called a decimal system

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Rationalize the denominator of 5/√3+root 2

goldenfox [79]

Answer:

10.075

Step-by-step explanation:

4 0

3 years ago

Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls

murzikaleks [220]

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

$P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}$

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= $(\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111$

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

$= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222$

Hence,

$P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684$

Part b

$P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}$

Cases for denominator when:

P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)

$= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}$

Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

$= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}$

Hence,

$P ( B = 12 / A+B+C = 21) = \frac{\frac{1}{144} }{\frac{1}{96} }\\\\= \frac{2}{3}$

4 0

3 years ago