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lutik1710 [3]
3 years ago
10

Please help need this done now hurry 1 question below!! 25 points

Mathematics
1 answer:
timofeeve [1]3 years ago
5 0
So we need to find the inequality, Simple! 
So we only have 5 dollars. That we cannot change. 
But the snadwich, however, we can change how much we buy. But this problem is asking us how many we can buy. So that is an x. So far, we have 2.25x.
Then we have to use the inequality sign 'less than or equal to' because we can buy less or equal to 5 dollars. 

So the equation is 2.25x \leq 5
Then we solve by isolating the x. We divide 2.25 both sides then we get x \leq 2
You might be interested in
|a+5|=−a−5, and can you solve |a+5|=a+5 please?
garik1379 [7]

Answer:

a1= -19, a2= 9

Step-by-step explanation:

step 1. Add 5 + 9 /a +5/ = 9+5

step 2. Use the absolute value definition, rewrite the absolute value equation as two separate equations. a+5=14 a+5= -14

step 3. a=9 a+5= -14

step 4. a= 9 a= -19 the equation has 2 solutions

a¹ = - 19, a²= 9

7 0
3 years ago
Help me quick pls 15 pts​
3241004551 [841]

Answer:

(-4, 4)

Step-by-step explanation:

The solution of f(x) - g(x) = 0 is the same as that of f(x) = g(x).

If f(x) = g(x), that implies that the two graphs cross one another.

The point of intersection is (-4, 4).  This is the solution.

5 0
3 years ago
Help me with this question
timama [110]

Answer:

15 square units

Step-by-step explanation:

5*3=15

5 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
4 years ago
(25 Points) The measure of b is ______?
Eva8 [605]

Answer:

21

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2 +b^2 = c^2

where a and b are the legs and c is the hypotenuse (opposite the right angle)

20^2 +b^2 = 29^2

400 + b^2 =841

Subtract 400 from each side

400-400 +b^2 = 841-400

b^2 = 441

Take the square root of each side

sqrt(b^2) = sqrt(441)

b = 21

6 0
3 years ago
Read 2 more answers
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