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kkurt [141]
3 years ago
13

There are (53)2 ⋅ 50 hens in a bird enclosure. What is the total number of hens in the enclosure?

Mathematics
2 answers:
Lisa [10]3 years ago
3 0
I got 5,300. I’m not sure if that’s what you meant for “d”.
topjm [15]3 years ago
3 0

Answer:

c) 5^{6}

Step-by-step explanation:

We have been given that there are (5^3)^2\cdot 5^0 hens in a bird enclosure. We are asked to find the total number of hens in the enclosure.

Using exponent property a^0=1, we will get:

(5^3)^2\cdot 1

(5^3)^2

Using exponent property (a^b)^c=a^{b\cdot c}, we will get:

5^{3\cdot 2}

5^{6}

Therefore, there are total number of hens in the enclosure would be 5^{6}.

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Stacy started a banking account with $150 and is spending $7 per day on lunch. This situation models which type of function?
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Linear

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What is the solution to the system of equations? 7x-2y= -24 5x+5y=15
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There are 10 true-false questions and 20 multiple choice questions from which to choose a five-question quiz how many ways can t
beks73 [17]

Answer:

In 68229 ways can the quiz be selected such that there is atleast three multiple choice questions

Step-by-step explanation:

Given:

Number of True or false questions= 10

Number of multiple choice questions= 20

To Find:

How many ways can 5 questions can be selected if there must be at least three multiple choice questions =?

Solution:

Combination

A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.  

The question States there sholud be ATLEAST 3 multiple choice question,

So, we may have

(3 Multiple choice question and 2 true or false question) or

(4 Multiple choice question and 1 true or false question) or

(5 Multiple choice question and 0 true or false question)

Required Number of ways = (20C3 X10C2) +(20C4 X10C1) + (20C5 X10C0)

Required Number of ways =(\frac{20!}{20!(20-3)!}\times\frac{10!}{10!(10-2)!})+(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-1)!}) +(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-0)!})

Required Number of ways = ( 1140 x 42) + (4845 x 10) +(15504 x 1)

Required Number of ways = 47880+48450+15504

Required Number of ways = 68229

3 0
2 years ago
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