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3 years ago

I need help with this plzz

Mathematics

2 answers:

Schach [20]3 years ago

6 0

Answer:

5/2

Step-by-step explanation:

rewona [7]3 years ago

6 0

Here you go! hope this helps

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a certain pharmaceutical company know that, on average 4% of a certain type of pill has an ingredient that is below the minimum

FinnZ [79.3K]

Using the binomial distribution, it is found that there is a 0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

For each pill, there are only two possible outcomes, either it is acceptable, or it is not. The probability of a pill being acceptable is independent of any other pill, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

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.

In this problem:

The sample has 20 pills, hence $n = 20$ .

100 - 4 = 96% are acceptable, hence $p = 0.96$

The probability that <u>fewer that 5 in a sample of 20 pills</u> will be acceptable is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

In which

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

$P(X = 0) = C_{20,0}.(0.96)^{0}.(0.04)^{20} = 0$

$P(X = 1) = C_{20,1}.(0.96)^{1}.(0.04)^{19} = 0$

$P(X = 2) = C_{20,2}.(0.96)^{2}.(0.04)^{18} = 0$

$P(X = 3) = C_{20,3}.(0.96)^{3}.(0.04)^{17} = 0$

$P(X = 4) = C_{20,4}.(0.96)^{4}.(0.04)^{16} = 0$

0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

A similar problem is given at brainly.com/question/24863377

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Jamel is asked to create triangles using three of four given sticks. The sticks measure 3 in., 6 in., 7 in., and 8 in. He create

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What is the surface area of a cylinder with a radius of 3ft and a height of 5ft

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3 years ago

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Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$

We get the $a^6b^2$ term when $k=2$ :

$k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2$

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1 year ago