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2 years ago

Erica has twice as many magazines as puzzle games. She has 30 of both in total. Write an

Mathematics

2 answers:

MArishka [77]2 years ago

5 0

She has 20 magazines and 10 puzzle games

m=magazines (20)
p=puzzle games (10)

m + p=30
m =2p
2p+p=30

kicyunya [14]2 years ago

4 0

Answer:

Well go ask Erica how many magazines she has.

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Help me plz I need it I'm very desperate help

Klio2033 [76]

Answer:

1/45

Step-by-step explanation:

There are 10 marbles

P ( yellow) = yellow marbles / total = 2/10 = 1/5

We do not replace the marble

There are 9 marbles, 1 of which is yellow

P ( yellow) = yellow marbles / total = 1/9

P ( yellow, no replacement, yellow) = 1/5 * 1/9 = 1/45

8 0

3 years ago

Read 2 more answers

Factor 20x2+25x-12x-15 by grouping

Hunter-Best [27]

Answer:

385-13x

Step-by-step explanation:

that is my answer

5 0

3 years ago

PLZZZZZZZZZZZZZZZZ HELP

Slav-nsk [51]

Answer:

The first box in the girl row is 18, the second one in the boy row is 8, and the bottom one is also 18.

Step-by-step explanation:

To get 81 to 9, you have to divide by 9, which means you have to do the same to 72. You get the ration 9 to 8, 9 being girls and 8 being guys. Then to get from 8 guys to 16, you have to multiply by 2, which means you do the same to 9 and get the answer of 18 girls. Hope this helps :)

5 0

3 years ago

I need help with both questions...

NikAS [45]

Answer:

17. a)

Step-by-step explanation:

17.

f(x) =x/2, then f(x-3)=(x-3)/2

x=3, f(x-3)=f(3-3)=f(0)=0 (values in table)

f(x-3)+3=0+3=3

18.

f(x) =x/2

f(x-2) for x=5 is f(5-2)=f(3)=3/2

f(x-2)+5=f(3)+5=3/2+5=3/2+10/2=13/2

3 0

3 years ago

Read 2 more answers

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago