All categories

3 years ago

This graph shows the solutions to the inequities y > - 1/3x + 5 and y < - 1/3x - 1.

Mathematics

2 answers:

Rus_ich [418]3 years ago

8 0

Answer: There is no solution

Step-by-step explanation:

on apex i got it wrong and it shows up as no solution so lol have fun

yaroslaw [1]3 years ago

6 0

It would be choice b

You might be interested in

A coin toss is used to determine which team will receive the ball at the beginning of a football game. The Cougars always choose

Delvig [45]

Yes, depending on part of the wind flow maybe. But the Cougars would win atleast 2/3 games

3 0

3 years ago

After a tropical storm in a certain state, news reports indicated that 19 percent of households in the state lost power during t

olya-2409 [2.1K]

Answer:

H0:p=0.19

Ha:p>0.19

Step-by-step explanation:

7 0

3 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.

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

5 0

3 years ago

10) What is the area of the figure below? Round your answer to the nearest tenth (1 place

Furkat [3]

75 units squared. separate the shapes into a rectangle and triangle, find the area of each, and add together

3 0

3 years ago

Unit cube:

Nutka1998 [239]

Answer:

12 unit cubes

Step-by-step explanation:

There are 4 unit cubes on each layer, and there are 3 layers as you can see. So 4 × 3 would equal 12.

8 0

3 years ago