6(0.25n−2)=n−0.5(5−2n)

Which graph represents the function? f(x) = ∛x+2

Gnoma [55]

Answer:

graph shown in image below

Step-by-step explanation:

4 0

3 years ago

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2x + 7 + 5x = 7 ( x + 8) how many solutions?

DanielleElmas [232]

Answer:

no solution

Step-by-step explanation:

5 0

3 years ago

recent survey found that 71% of all adults over 50 wear glasses for driving. In a random sample of 30 adults over 50, what is

valentina_108 [34]

Answer:

Mean = 21.3

Standard Deviation = 2.48

Step-by-step explanation:

We are given the following in the question:

p = 71% = 0.71

Sample size, n = 30

We have to find the mean and the standard deviation of those that wear glasses.

$\text{Mean} = np\\=30(0.71)\\=21.3$

Thus, the mean of those who wear glasses is 21.3

$\text{Standard deviation} = \sqrt{npq}\\=\sqrt{np(1-p)}\\=\sqrt{30\times 0.71\times (1-0.71)}\\=2.48$

Thus, the standard deviation of those who wear glasses is 2.48

5 0

3 years ago

Givenf (x) = { (+ 6) whatisf (-6)?

zaharov [31]

Answer:

$f(x) = 6 \\ f( - 6) = 6 \\ because \: here \: function \: is \: \\ independent \: of \: x \\ thank \: you$

4 0

3 years ago

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Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list

Radda [10]

Answer:

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: Any from 0 to 5.

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this question:

$n = 5, p = 0.3, q = 1 - p = 0.7$

So

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

8 0

3 years ago