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Ber [7]
3 years ago
11

Need help by thx iejhbeeje​

Mathematics
2 answers:
jek_recluse [69]3 years ago
4 0

Answer:

$14.94

Step-by-step explanation:

$40-$10.02= $29.98

2÷29.98= $14.94

Elina [12.6K]3 years ago
3 0

Answer:

Step-by-step explanation:bhj

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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

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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
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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 <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> 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.

Consider homogeneous equation y''-2y'=0

Let y=e^{rt} be the solution .

We get (r^2-2r)e^{rt}=0

Since e^{rt}\neq 0, r^2-2r=0.

So, we get solution as y_c=c_1+c_2e^{2t}

As constant term and e^{2t} are already in the R.H.S of equation

y" - 2y' = 6t + 5e^{2t}, we can take y_p as y_p=A+Bt+Ce^{2t}

6 0
3 years ago
What is 83% as a decimal SHOW WORK IF U CAN HELP ASAP!!
Marat540 [252]

0.83Answer:

Step-by-step explanation:

83% means 83 per 100 or 83/100 if you divide 83 by 100 you get 0.83

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