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Which quadrant will (30,-45) fall into?

Varvara68 [4.7K]

Answer:

Quadrant 4

Step-by-step explanation:

The x value is positive so it is either in 1 or 4

The y value is negative so it is either 3 or 4

To meet both conditions, it must be 4

6 0

2 years ago

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Mr. Mosallam invested $14,000 in equipment to print yearbooks for Fordson High School. Each yearbook costs $7 to print and sells

earnstyle [38]

He needs to sell 500 year books to break even

5 0

3 years ago

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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.

(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)

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

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

Given: y" - 2y' = 6t + 5e^2t. Find the correct form to use for y_p if the equation is solved using Undetermined coefficients. Do

const2013 [10]

Answer:

$y_p=A+Bt+Ce^{2t}$

Step-by-step explanation:

Given: $y'' - 2y' = 6t + 5e^{2t}$ .

we need to find the correct form for $y_p$ if the equation is solve using undetermined coefficients.

A first order differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}=f\left ( x,y \right )$ is said to be homogeneous if $f(tx,ty)=f(x,y)$ for all t.