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sergey [27]
2 years ago
6

Please I need help with this and show your steps

Mathematics
2 answers:
Sidana [21]2 years ago
5 0

Answer:

1

Step-by-step explanation:

2(5)-(1+2)^2

multiply: 10-(1+2)^2

add: 10-(3)^2

square (multiply 3 by itself): 10-9=1

krek1111 [17]2 years ago
3 0
Here you go! Hope this helps!

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There are 16 females and 20 males in a class.
ozzi
20/36, or 5/9 if simplified.
5 0
3 years ago
Read 2 more answers
What is the value of y?
solong [7]
Sum of interior angles in a triangle = 180
so
2y + y+10 + 50 = 180
3y + 60 = 180
3y = 120
  y = 40

Answer is A. 40
8 0
3 years ago
Chi has $11:30 in dimes and quarters. The number of dimes is three more
Oliga [24]

Answer:

To solve this question, we first have to know that we are trying to assemble

and solve equations with two variables: x

and y.

Assuming x

is the number of dimes and y

is the number of quarters, we

can assemble the equations.

since the number of dimes is 3 more than three times the number of quarters,

we get x=3y+3.

Additionally, multiplying the number of dimes by 0.1 and the number of

quarters by 0.25,

we get the total value of x

and y

which appears by this

equation 0.1x+0.25y=11.30

.

Substituting the value of x

in the second equation, we get 0.1(3y+3)+

0.25y=11.30

.

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we get x=3×20+3=

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Chapter 9

Math Models and Geometry

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Prealgebra

Step-by-step explanation:

6 0
3 years ago
Let X1, . . . ,Xn ∈ R be independent random variables with a common CDF F0. Let Fn be their ECDF and let F be any CDF. If F = Fn
Georgia [21]

Answer:

See the proof below.

Step-by-step explanation:

For this case we need to proof that: Let X_1, X_2, ...X_n \in R be independent random variables with a common CDF F_0. Let F_n be their ECDF and let F any CDF. If F \neq F_n then L(F)

Proof

Let z_a different values in the set {X_1,X_2,...,X_n}} and we can assume that n_j \geq 1 represent the number of X_i that are equal to z_j.

We can define p_j = F(z_j) +F(z_j-) and assuming the probability \hat p_j = \frac{n_j}{n}.

For the case when p_j =0 for any j=1,....,m then we have that the L(F) =0< L(F_n)

And for the case when all p_j >0 and for at least one p_j \neq \hat p_j we know that log(x) \leq x-1 for all the possible values x>0. So then we can define the following ratio like this:

log (\frac{L(F)}{L(F_n)}) = \sum_{j=1}^m n_j log (\frac{p_j}{\hat p_j})

log (\frac{L(F)}{L(F_n)}) = n \sum_{j=1}^m \hat p_j log(\frac{p_j}{\hat p_j})

log (\frac{L(F)}{L(F_n)}) < n\sum_{j=1}^m \hat p_j (\frac{p_j}{\hat p_j} -1)

So then we have that:

log (\frac{L(F)}{L(F_n)}) \leq 0

And the log for a number is 0 or negative when the number is between 0 and 1, so then on this case we can ensure that L(F) \leq L(F_n)

And with that we complete the proof.

8 0
3 years ago
Solve the following equation. –4y + 8 = 4(2y – 2) – 2(–16 + 8y) A. y = –4 B. y = 4 C. y = –2 D. y = –6 I need Help I am utterly
leonid [27]
The answer is B. (y = 4)


Further explanation:

–4y + 8 = 4(2y – 2) – 2(–16 + 8y)
= -4y + 8 = 4 * (2y -2) - 2 * (-16 + 8y)
= -4y + 8 = ( 4 * 2y - 4 *2 ) - (2 * -16 - 2 * 8y )
= -4y + 8 = 8y - 8 - ( -32 - 16y)
= -4y + 8 = 8y - 8 + 32 + 16y
Move y’s to the left & numbers to the right
= -4y - 8y - 16y = -8 + 32 - 8
= 12y - 16y = -16 + 32
Y = 16 / 28
Y = 4 = (B choice)
3 0
3 years ago
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