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3 years ago

Zara bought 150 pairs of shoes

Mathematics

2 answers:

Bad White [126]3 years ago

6 0

Zara profited 1750 dollars

Leto [7]3 years ago

3 0

required profit=$1750

cp=$6000

sp1=100×$50=$5000

sp of remaining sp2={150-100}×$55=50×55=$2750

profit={sp1+sp2}-cp=$5000+$2750-$6000=$1750

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1. Howard's CD House sells new CDs for $15 each and used CDs for $6 each. Bre'anna bought 51 CDs and spent a

Yanka [14]

The numbers of new CDs is 30 and the old CDs is 21.

To solve this problem, we have to write a system of equations in which we would know the numbers of each types of CD she bought.

<h3>System of Equations</h3>

This is used to solve a series or complex of word problems in which we can represent in a mathematical statements;

Let n represents numbers of new CDs
Let u represents numbers of old CDs

The equations can be represented as

$n + u = 51 ...equation(i)\\15n + 6u = 576 ... equation(ii)$

Let's take equation (i)

Making n the subject of formula

$n + u = 51\\n = 51 - u...equation(iii)$

we can put equation (iii) into equation (ii)

$15n + 6u =576\\n = 51 - u\\15(51 - u) + 6u = 576\\765-15u + 6u = 576\\collect like terms\\765-576=15u - 6u\\189=9u\\9u/9 = 189/9\\u = 21$

Let's put the value of u into equation (i)

$n + u = 51\\u = 21\\n + 21 = 51\\n = 51 - 21\\n = 30$

From the calculations above, the numbers of new CDs is 30 and the old CDs is 21.

Learn more on system of equations here;

brainly.com/question/13729904

4 0

2 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$