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Andreyy89
2 years ago
5

What is 5 11/10 simplified

Mathematics
1 answer:
lbvjy [14]2 years ago
3 0

Answer:

61/10 or in decimal form 6.1 or 6 1/10

Step-by-step explanation:

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Each person in a group of college students was identified by graduating year and asked when he or she preferred taking classes:
swat32

Answer:

<em>0.615</em>

Step-by-step explanation:

The frequency table is attached below.

We have to calculate, the probability that the student preferred morning classes given he or she is a junior.

i.e P(\text{Morning }|\text{ Junior})

We know that,

P(A\ |\ B)=\dfrac{P(A\ \cap\ B)}{P(B)}

So,

P(\text{Morning }|\text{ Junior})=\dfrac{P(\text{Morning }\cap \text{ Junior})}{P(\text{Junior})}

Putting the values from the table,

\dfrac{P(\text{Morning }\cap \text{ Junior})}{P(\text{Junior})}=\dfrac{\frac{16}{143}}{\frac{26}{143}}=\dfrac{16}{26}=0.615

3 0
3 years ago
Write an equation for each sentence and then solve.
Whitepunk [10]

Answer:

Step-by-step explanation:

Let the two consecutive numbers be 2n and 2n+2

The sum of these numbers in 2n + (2n+2).

Hence

2n + (2n+2) = 34

4n + 2 = 34

4n  = 32

n =8.

8 0
2 years ago
Show me how to solve .384-.3798
-Dominant- [34]

Answer:

0.0042

Step-by-step explanation:

0.3840

- 0.3798

‐--------------

0.0042

‐--------------

8 0
3 years ago
The cost of renting an apartment in the downtown city area is increasing by 4% each year. Currently, the average rent for
Novay_Z [31]

Answer:500

Step-by-step explanation:

4 0
2 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.

3 0
3 years ago
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