-2/5x + 2 find the inverse

3. Assume that the likelihood of a child under the age of ten watching PBS is 0.76. Three children are

natta225 [31]

Using the binomial distribution, the probabilities are given as follows:

a) 0.4159 = 41.59%.

b) 0.5610 = 56.10%.

c) 0.8549 = 85.49%.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

For this problem, the values of the parameters are:

n = 3, p = 0.76.

Item a:

The probability is P(X = 2), hence:

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

$P(X = 2) = C_{3,2}.(0.76)^{2}.(0.24)^{1} = 0.4159$

Item b:

The probability is P(X < 3), hence:

P(X < 3) = 1 - P(X = 3)

In which:

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

$P(X = 3) = C_{3,3}.(0.76)^{3}.(0.24)^{0} = 0.4390$

Then:

P(X < 3) = 1 - P(X = 3) = 1 - 0.4390 = 0.5610 = 56.10%.

Item c:

The probability is:

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4159 + 0.4390 = 0.8549$

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

4 0

2 years ago

I need this done today whoever gets this correct gets brainliest!!! PLZZ HELPP MEE!!!<br> hurryy!!

kenny6666 [7]

Answer:

i dont know if this is right but 0.125 for every one cup i got this by dividing the fractions

3 0

3 years ago

Evaluate the expression below at x=5

Citrus2011 [14]

5 is the answer to the promblem

5 0

3 years ago

With the holidays coming I have set a budget. I have $600 to spend. After I buy Nory and Nolan's gift, I have $350 left. What pe

ahrayia [7]

You have 250 left in your budget

4 0

3 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

5 0

3 years ago