The table shows the outputs y for different inputs x:

Mathematics

2 answers:

andre [41]3 years ago

8 0

Part A: Yes, because no input yields more than output values.

Part B: If f(x)= 5x-21, switch the x's to 11.

f(11)=5(11)- 21. 5 times 11 equals 55, then minus by 21. 55- 21= 34.

In the table, y=8 and x=11, but in the function, y=34 and x=11.

Part C: I have to make sure to plug in 99 for f(x).

f(x)= 5x -21, which basically is 99= 5x -21.

Add 21 to both sides. 99+21 = 5x -21+21.

99 plus 21 =120, which 120= 5x.

Divide 120 by 5 and divide 5x by 5.

The answer will be 24

Rasek [7]3 years ago

5 0

Part A: Yes, the data represent a function. The definition of a function is a relation in which no value of x will have two different values of y.
(Every time you plug in 3 as x, you will always get 4 as y; it's ok if you plug in 3 and 5 as x and get the same y, you just can't get two different y's for one x; sorry, it is pretty confusing). None of the numbers in the table repeat, so we can safely say that the relation is a function.

Part B: All we have to do is plug in 11 for x in the function given to find the answer:
$f(x)=5x-21\\f(x)=5(11)-21\\f(x)=55-21\\f(x)=34$
In the table, y = 8 when x = 11, but in the function given, y = 34 when x = 11, so the function given is greater.

Part C: To find the answer to C, just plug in 99 for f(x), as it tells you to do:
$f(x) = 5x-21\\99=5x-21\\99+21=5x-21+21\\120=5x\\120/5=5x/5\\24=x$

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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$