Is y=2^x linear, exponential or neither

1 answer:

3 0

Answer:
exponential

Step-by-step explanation:
An exponential function is defined as a function with a positive constant other than 1 raised to a variable exponent. A function is evaluated by solving at a specific input value. An exponential model can be found when the growth rate and initial value are known.

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Given the venn diagram, find P(brother and sister)

LUCKY_DIMON [66]

Based on the venn diagram of siblings, the probability of having a brother and sister is c. 0.33.

<h3>What is the probability of brother and sister?</h3>

The probability of having a brother and sister can be found as:

= Number of those with brother and sister / Total number of people in poll

Solving gives:

= 10 / (8 + 10 + 5 + 7)

= 10 / 30

= 0.33

In conclusion, the probability of having a brother and sister is 0.33.

Find out more on Venn diagrams at brainly.com/question/9085273

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6 0

2 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .