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Vedmedyk [2.9K]
3 years ago
7

Jane has six different hamsters for which she has two cages, one red and one blue. She wants to put three hamsters in the red ca

ge and three in the blue cage, but two of the hamsters, Felix and Oscar, do not get along and cannot be in the same cage. In how many different ways can she choose which three hamsters to put in the red cage?
Mathematics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

<u>12 ways</u>

Step-by-step explanation:

The different ways to choose which three hamsters to put in the red cage.

At first , she must choose one of the two of the hamsters, Felix and Oscar, because they do not get along and cannot be in the same cage.

So, the number of ways to choose one of them = 2

Then she will choose 2 from the remaining 4 hamsters (order does not matter)

The number of ways to choose 2 from 4 = 4C2 = 6

So, the total number of ways = 2 * 6 = 12

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The table when completed looks like this:

x   -2   -1   0   1   2

f(x) 4   2   1    2   4.

We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).

<h3>What is a function?</h3>

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We have been given a function,

$f(x)=\left \{ {{(1/2)^{x},x\leq 0 } \atop {2^x > 0}} \right.

We have to estimate the value of f(x) when x = {-2, -1, 0, 1, 2}

If the values of x ≤ 0, take f(x) = (1/2)ˣ then

f(-2) = (1/2)⁻² = 2² = 4

f(-1) = (1/2)⁻¹ = 2¹ = 2

f(0) = (1/2)⁰ = 1

If the values of x > 0,  take f(x) = 2ˣ then

f(1) = 2¹ = 2

f(2) = 2² = 4

The table exists as follows

x   -2   -1   0   1   2

f(x) 4   2   1    2   4.

We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).

To learn more about independent and dependent variable refer to:

brainly.com/question/11719274

#SPJ9

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1 year ago
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