All categories

4 years ago

I need help please !!!!

Mathematics

2 answers:

lora16 [44]4 years ago

4 0

$\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\sqrt{c^5}=\sqrt[2]{c^5}=c^\frac{5}{2}$

irga5000 [103]4 years ago

3 0

The square root of a number means you take that number to the 1/2 power. Our number is
c^5. Taking its square root, we have
(c^5)^(1/2). When you have exponents like that, you can multiply them together. For example,
(x^2)^3 = x^(2*3) = x^6

In this case, we have
(c^5)^(1/2) = c^(5*(1/2)) = c^(5/2)

You might be interested in

PLEASE HELP<br> Identify the first five terms of the sequence in which a, = 3n2 - 1.

Neko [114]

Step-by-step explanation:

you cannot just put the actual numbers in and calculate ?

and you can't provide the correct problem statement, as it seems.

I assume you mean

an = 3n² - 1

a sequence starts with a1, so, n>=1

a1 = 3×1² - 1 = 3-1 = 2

a2 = 3×2² -1 = 3×4 - 1 = 12 - 1 = 11

a3 = 3×3² - 1 = 3×9 - 1 = 27 - 1 = 26

a4 = 3×4² - 1 = 3×16 - 1 = 48 - 1 = 47

a5 = 3×5² - 1 = 3×25 - 1 = 75 - 1 = 74

there, that is all there is to it. you really needed help with that ?

6 0

3 years ago

105. Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independe

Ainat [17]

Answer:

a. X is the number of adults in America that need to be surveyed until finding the first one that will watch the Super Bowl.

b. X can take any integer that is greater than or equal to 1. $\rm X\in \mathbb{Z}^{+}$ .

c. $\rm X \sim NB(1, 0.40)$ .

d. $E(\rm X) = 2.5$ .

e. $P(\rm X = 7) = 0.0187$ .

f. $P(\text{X} = 3) +P(\text{X} = 4) = 0.230$ .

Step-by-step explanation:

In this setting, finding an adult in America that will watch the Super Bowl is a success. The question assumes that the chance of success is constant for each trial. The question is interested in the number of trials before the first success. Let X be the number of adults in America that needs to be surveyed until finding the first one who will watch the Super Bowl.

It takes at least one trial to find the first success. However, there's rare opportunity that it might take infinitely many trials. Thus, X may take any integer value that is greater than or equal to one. In other words, X can be any positive integer: $\rm X\in \mathbb{Z}^{+}$ .

There are two discrete distributions that may model X:

The geometric distribution. A geometric random variable measures the number of trials before the first success. This distribution takes only one parameter: the chance of success on each trial.
The negative binomial distribution. A negative binomial random variable measures the number of trials before the r-th success. This distribution takes two parameters: the number of successes $r$ and the chance of success on each trial $p$ .

$\rm NB(1, p)$ (note that $r=1$ ) is equivalent to $\sim Geo(p)$ . However, in this question the distribution of $\rm X$ takes two parameters, which implies that $\rm X$ shall follow the negative binomial distribution rather than the geometric distribution. The probability of success on each trial is $40\% = 0.40$ .

$\rm X\sim NB(1, 0.40)$ .

The expected value of a negative binomial random variable is equal to the number of required successes over the chance of success on each trial. In other words,

$\displaystyle E(\text{X}) = \frac{r}{p} = \frac{1}{0.40} = 2.5$ .

$P(\rm X = 7) = 0.0187$ .

Some calculators do not come with support for the negative binomial distribution. There's a walkaround for that as long as the calculator supports the binomial distribution. The r-th success occurs on the n-th trial translates to (r-1) successes on the first (n-1) trials, plus another success on the n-th trial. Find the chance of (r-1) successes in the first (n-1) trials and multiply that with the chance of success on the n-th trial.

$P(\text{X} = 3)+P(\text{X} = 4) = 0.230$ .