All categories

3 years ago

What is the total amount 2 5/8 + 1 3/4 =

Mathematics

1 answer:

trasher [3.6K]3 years ago

8 0

Answer:4 3/8

Step-by-step explanation: you add the full numbers together then you have to change the denominator of 3/4 to be /8 and what you do to the bottom you have to do to the top then you get 3 11/8 you then see how many 8 go into 11 to se if you need to add another full number and the numbers left over go back to the 8 making the awnser 4 and 3/8.

You might be interested in

You and your business partner are mailing advertising flyers to your customers. You address 6 flyers

jonny [76]

Answer:

140

Step-by-step explanation:

3 0

3 years ago

Find the slope intercept form of the line through (3,4) and (-2,4)

Alchen [17]

Answer:

y = 4

Step-by-step explanation:

As we move from the point (-2, 4) to the point (3, 4), x increases by 5 units, but y stays the same. We know immediately that the slope, m = rise / run, is zero, because of this.

Both points are on the horizontal line y = 4. This answer is sufficient in itself, but could be re-written as y = 0x + 4.

5 0

3 years ago

Which option is correct for finding x?

san4es73 [151]

Answer:

35 degrees is correct(I might be wrong though.)

Step-by-step explanation:

Since this is a vertical angle question,3x° = 70 + x°

Therefore,3x - x = 2 x,making 2x=70.

so x= 70÷ 2=35. Again- I might be wrong-

5 0

3 years ago

Somebody please assist me here

Anettt [7]

The base case of $n=1$ is trivially true, since

$\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)$

but I think the case of $n=2$ may be a bit more convincing in this role. We have by the inclusion/exclusion principle

$\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)$

with equality if $E_1\cap E_2=\emptyset$ .

Now assume the case of $n=k$ is true, that

$\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)$

We want to use this to prove the claim for $n=k+1$ , that

$\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$

The I/EP tells us

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)$

and by the same argument as in the $n=2$ case, this leads to

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})$

By the induction hypothesis, we have an upper bound for the probability of the union of the $E_1$ through $E_k$ . The result follows.

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$

5 0

2 years ago

Simplify<br> 2×7p-3×4q+15p/3-24q/6

Lelechka [254]

Answer:

ans=3

Step-by-step explanation:

5 0

3 years ago