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Dmitriy789 [7]
3 years ago
5

I am between 7,000,000 and 8,000,000. all my digits are odd. all the digits in my thousands period are the same. the sum of my d

igit is 31. what number am i? give as many answers as you can. what strategies did you use to find the mystrey number
Mathematics
2 answers:
lozanna [386]3 years ago
7 0

Answer:

The number can be many numbers like: 7,777,111  or 7,555,333  or 7,111,777

A: 7,777,111

We have four 7's and three 1's

7*4=28

1*3=3

28+3=31

B: 7,555,333

We have one 7, three 5's and three 3's.

5*3=15

3*3=9

15+9+7=31

These also can be 7,333,555 or 7,111,777.

aniked [119]3 years ago
4 0
There are many possibilities so I am going to give 3 answers 777113 777131 777311
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3 years ago
A croissant shop produces two products: bear claws (B) and almond filled croissants (C). Each bear claw requires 6 oz of flour,
Kruka [31]

Answer:

letter A: B = 400; C = 1000; Max Z = $380

Step-by-step explanation:

bear claws (B) need 6 oz of flour (f), 1 oz of yeast (y), 2 ts of paste (p)

croissants (C) need 3 oz of flour (f), 1 oz of yeast (y), 4 ts of paste (p)

putting that information in equations, we have:

B = 6f + y + 2p

C = 3f + y + 4p

The total number of resources (R) are:

R = 6600f + 1400y + 4800p

Let's call M our total profit, "k1" the number of B produced and "k2" the number of C produced.

So, we can state that:

M = k1*0.2 + k2*0.3

The number of resources R2 will demand is calculated like this:

R2 = k1*B + k2*C = (6k1+3k2)f + (k1+k2)y + (2k1+4k2)p

using R2 and R, we can make some inequations:

6k1+3k2 <= 6600 -> 2k1+k2 <= 2200

k1+k2 <= 1400

2k1+4k2 <= 4800 -> k1+2k2 <= 2400

if we try to maximize k2 (as it worths more), we will have k1 = 0 and k2 = 1200 (limited by p), but looking at the resources R, we will still have resources to use (f and y). Looking at B and C expressions, we see that removing one C gives enough 'p' to make 2 B, which is a good trade (as 2B worths 0.4 cents, and 1C worths 0.3 cents). we have 200 'y' remaining, so doing this 200 times give us k1 = 400 and k2 = 1000, and the only resource remaining will be some of 'f'.

calculating the profit M, we have:

M = 400*0.2 + 1000*0.3 = 380$

the right answer is letter A.

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3 years ago
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o-na [289]

Answer:

  • <u><em>About 0.22</em></u>

Explanation:

There are two sets:

  • Set W of incoming seniors who took AP World History, and
  • Set E of incoming seniors who took AP European History

And there is a subset, which is the intersection of those two sets:

  • Subset W ∩ E of senior students who took both.

The incoming seniors who are allowed to enroll in AP U.S. History, call them the subset S, is the set of those students that belong to W or E or both W ∩E.

By property of sets:

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Then, 178 out of 825 incoming seniors took one or both courses, and the desired probability of a randomly selected incoming senior is allowed to enroll in AP U.S. History is:

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