I stink at math how about some smart people check out my page and anwser some questions I give brainiest for correct

Mathematics

1 answer:

ziro4ka [17]3 years ago

4 0

The 11% seems correct by my calculations but you forgot the decimal in it. The correct answer is .11% I believe

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A machine making computer chips isn't working correctly, and 5% of the computer chips it makes are defective. If an inspector ch

Rainbow [258]

Answer:

0.25% probability that they are both defective

Step-by-step explanation:

For each computer chip, there are only two possible outcomes. Either they are defective, or they are not. The probability of a computer chip being defective is independent of other chips. So we use the binomial probability distribution to solve this question.

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.

5% of the computer chips it makes are defective.

This means that $p = 0.05$

If an inspector chooses two computer chips randomly (meaning they are independent from each other), what is the probability that they are both defective?

This is P(X = 2) when n = 2. So

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

$P(X = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025$

0.25% probability that they are both defective

4 0

3 years ago

Evaluate the Riemann sum for (x) = x3 − 6x, for 0 ≤ x ≤ 3 with six subintervals, taking the sample points, xi, to be the right e

DENIUS [597]

Hello,

Reminders:
$$\sum_{i=1}^{n} i= \dfrac{n(n+1)}{2}$


$
$$ \sum_{i=1}^{n} i^2= \dfrac{n(n+1)}{2}$

$
$$ \sum_{i=1}^{n} i^3= \dfrac{n^2(n+1)^2}{4} $
$

$n=6\\
x_{0}=0\\
x_{n}=3\\
\Delta= \dfrac{x_n-x_0}{n}=0.5\\

$
$x_{i} =0+\Delta*i= \frac{i}{2}$

$\sum_{i=1}^{n}\ \Delta*f(x_{i})=\sum_{i=1}^{6}\ \dfrac{i^3/8-6i}{2}$
$$= \dfrac{1}{2}\sum_{i=1}^{6} (\frac{i^3}{8} -6i)=\dfrac{1}{16}*\frac{(6*7)^2}{4}- \frac{9*7}{2}= -3.9575$

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