Define a variable , write an inequality , and solve the problem:

Please help extra points and Mark brainless

Naya [18.7K]

-1/6 Is the answer for this!!

4 0

3 years ago

Please help I don't understand. file uploaded.

IrinaK [193]

Answer:

ugsrgnSRernynerytydheyhxeryfrxjvvvbhfrea3sxrggdho

8 0

3 years ago

A rectangular auditorium seats 1564 people. the number of seats in each row exceeds the number of rows by 1212. find the number

yanalaym [24]

We can safely assume that 1212 is a misprint and the number of seats in a row exceeds the number of rows by 12. Let r = # of rows and s = # of seats in a row. Then, the total # of seats is T = r x s = r x ( r + 12), since s is 12 more than the # of rows. Then r x (r + 12) = 1564 or r**2 + 12*r - 1564 = 0, which is a quadratic equation. The general solution of a quadratic equation is: x = (-b +or- square-root( b**2 - 4ac))/2a In our case, a = 1, b = +12 and c = -1564, so x = (-12 +or- square-root( 12*12 - 4*1*(-1564) ) ) / 2*1 = (-12 +or- square-root( 144 + 6256 ) ) / 2 = (-12 +or- square-root( 6400 ) ) / 2 = (-12 +or- 80) / 2 = 34 or - 46 We ignore -46 since negative rows are not possible, and have: rows = 34 and seats per row = 34 + 12 = 46 as a check 34 x 46 = 1564 = total seats

4 0

3 years ago

Kabul’s bookshop marks up all books by 40 percent of their cost. The overhead rate is 16 percent of the selling price. What is t

morpeh [17]

Answer:

The net profit rate on the book is 5.41

Step-by-step explanation:

Given as :

The marks up percentage of book = m = 40%

The overhead rate is 16% of selling price

The cost price of book = c.p = $18.10

Let The profit = $p

Let The selling price = s.p

Now, According to question

mark up percentage = $\dfrac{s.p - c.p}{c.p}$

I.e 40% = $\dfrac{s.p - 18.10}{18.10}$

Or, $\dfrac{40}{100}$ + 1 = $\dfrac{s.p}{18.10}$

Or, $\dfrac{140}{100}$ = $\dfrac{s.p}{18.10}$

Or, s. p = $\dfrac{140\times 18.10}{100}$

∴ s.p = $25.34

So, selling price of book = s.p = $25.34

Now, The overhead percentage = 16%

i.e overhead rate = $\dfrac{\textrm estimated cost}{\textrm estimated total base unit}$

Or, estimated cost = 16% × 25.34

I,e estimated cost = 0.16 × 25.34

∴ estimated cost = $4.05

Now,

Profit = selling price of book - estimated book cost

I.e p = $25.34 - $4.05

∴ p = $21.29

So, The profit rate% = $\dfrac{\textrm profit}{\textrm estimated cost}$

I.e The profit rate% = $\dfrac{21.29}{4.05}$

∴ profit rate %= 5.41

So, The profit rate = p = 5.41

Hence, The net profit rate on the book is 5.41 Answer

4 0

3 years ago

The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab

kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable X can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable X thus follows a Poisson distribution with parameter, λ = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.

The mean of the approximated distribution of X is:

μ = λ = 250

The standard deviation of the approximated distribution of X is:

σ = √λ = √250 = 15.8114

Thus, $X\sim N(250, 250)$

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

$P(235$

$=P(-0.95$

Thus, the value of P (235 < X < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

$P(48$

$=P(-0.28$

Thus, the value of $P(48$ .

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

$P(48$

$=P(-0.40$

Thus, the value of $P(48$ .

(d)

Let the sample size be n.

$P(48$

$0.95=P(-z$

The value of z for this probability is,

z = 1.96

Compute the value of n as follows:

$z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241$

Thus, the sample selected must be of size 240.

5 0

3 years ago