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2 years ago

Name the object that exhibits rotational symmetry.

Mathematics

1 answer:

salantis [7]2 years ago

8 0

Answer:

circular plate

Step-by-step explanation:

Rotational symmetry is attained on a figure if it is possible to achieve the same image of the figure after rotating it. For example , rotating a circular plate marked with numbers 1 to 6 through a 360° rotation, the number marked 1 with attain a position similar like that before rotation.

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A pharmaceutical company receives large shipments of ibuprofen tablets and uses an acceptance sampling plan. This plan randomly

Keith_Richards [23]

Answer:

0.7208 = 72.08% probability that this whole shipment will be accepted.

Step-by-step explanation:

For each tablet, there are only two possible outcomes. Either it meets the required specifications, or it does not. The probability of a tablet meeting the required specifications is independent of any other tablet, which means that the binomial probability distribution is used 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.

4% rate of defects

This means that $p = 0.04$

26 tablets

This means that $n = 26$

What is the probability that this whole shipment will be accepted?

Probability that at most one tablet does not meet the specifications, which is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Thus

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

$P(X = 0) = C_{26,0}.(0.04)^{0}.(0.96)^{26} = 0.3460$

$P(X = 1) = C_{26,1}.(0.04)^{1}.(0.96)^{25} = 0.3748$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3460 + 0.3748 = 0.7208$

0.7208 = 72.08% probability that this whole shipment will be accepted.

3 0

3 years ago

Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).

ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

$\overrightarrow {PQ}==$

$\overrightarrow {PR}==$

The nonzero vector orthogonal to the plane through the points P,Q, and R is

$\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}$

$\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}$

Expand along row 1.

$\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}$