Sean bakes 18 dozen chocolate cupcakes and 13 dozen vanilla cupcakes. About how many cupcakes did Sean bake? EXPLAIN.

Evaluate −m2n+m2 when m=−4.

fgiga [73]

Answer:

Does this answer your question − 16 n + 16

4 0

3 years ago

Which graph is and is not a function!?

tresset_1 [31]

Answer:

The first one is a function. The other two are not functions.

Step-by-step explanation:

To find this, you use the vertical line test. If the vertical line touches multiple points on the graphed line, it is not a function. If the vertical line only touches the graphed line at one point, it is a function.

Good Luck! I hope this helps!

5 0

3 years ago

Determine x when y = 18, if y = 9 when x = 8

tensa zangetsu [6.8K]

Answer:x = 16

Step-by-step explanation: Make the equation 8/9=x/18: Cross multiply to get 9x=144 and divide both sides by 9 to get 16

Hope this helps, HAVE A BLESSED AND WONDERFUL DAY! As well as a great Superbowl Weekend! :-)

- Cutiepatutie ☺❀❤

4 0

3 years ago

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.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

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)

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