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Lelechka [254]
3 years ago
5

Write 343 with an exponent by using 7 as the base

Mathematics
1 answer:
inessss [21]3 years ago
8 0
If 7 is the base then the exponent is 3.
7x7x7=343

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Plz help very important question!!
Zepler [3.9K]

Answer:

the domain is: 0,-3,4,2,-2.

Step-by-step explanation:

6 0
3 years ago
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James is a welder who earns $20 per hour. He works 8 hours per day, Monday through Friday. How much does James earn in one week?
lara [203]
20×8=160
160×5=800
james earns $800 in one week.
3 0
4 years ago
Deluxe River Cruises operates a fleet of river vessels. The fleet has two types of vessels: A type-A vessel has 60 deluxe cabins
bixtya [17]

Answer: It should be used 2 for type-A and 3 for type-B to minimize the cost.

Step-by-step explanation: As it is stipulated, <u>x</u> relates to type-A and y to type-B.

Type-A has 60 deluxe cabins and B has 80. It is needed a minimum of 360 deluxe cabins, so:

60x + 80y ≤ 360

For the standard cabin, there are in A 160 and in B 120. The need is for 680, so:

160x + 120y ≤ 680

To calculate how many of each type you need:

60x + 80y ≤ 360

160x + 120y ≤ 680

Isolating x from the first equation:

x = \frac{360 - 80y}{60}

Substituing x into the second equation:

160(\frac{360 - 80y}{60}) + 120y = 680

-3200y+1800y = 10200 - 14400

1400y = 4200

y = 3

With y, find x:

x = \frac{360 - 80y}{60}

x = \frac{360 - 80.3}{60}

x = 2

To determine the cost:

cost = 42,000x + 51,000y

cost = 42000.2 + 51000.3

cost = 161400

To keep it in a minimun cost, it is needed 2 vessels of Type-A and 3 vessels of Type-B, to a cost of $161400

8 0
3 years ago
each of the 20 balls is tossed independently and at random into one of the 5 bins. let p be the probability that some bin ends u
amm1812

if p is the probability that some bin ends up with 3 balls and q is the probability that every bin ends up with 4 balls. pq is 16.

First, let us label the bins with 1,2,3,4,5.

Applying multinomial distribution with parameters  n=20  and  p1=p2=p3=p4=p5=15  we find that probability that bin1 ends up with 3, bin2 with 5 and bin3, bin4 and bin5 with 4 balls equals:

5−2020!3!5!4!4!4!

But of course, there are more possibilities for the same division  (3,5,4,4,4)  and to get the probability that one of the bins contains 3, another 5, et cetera we must multiply with the number of quintuples that has one 3, one 5, and three 4's. This leads to the following:

p=20×5−2020!3!5!4!4!4!

In a similar way we find:

q=1×5−2020!4!4!4!4!4!

So:

pq=20×4!4!4!4!4!3!5!4!4!4!=20×45=16

thus, pq = 16.

To learn more about Probability visit: brainly.com/question/29508225

#SPJ4

7 0
1 year ago
Help me please, not very good at math. Thank u:)
ozzi

Answer:

76cm^2

Step-by-step explanation:

Separate the shape into two different rectangles. Then add the area of the rectangles together.

4 x 4 = 16

6 x 10 = 60

16 + 60 = 76

8 0
3 years ago
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