Which linear inequality is represented by the graph? y ≥ 1/3x – 4 y ≤ 1/3x – 4 y ≤ 1/3x + 4 y ≥ 1/3x + 4

Mathematics

2 answers:

VikaD [51]3 years ago

7 0

Solid line...means equal sign is present.....shading below the line means less then....u have a y int at (0,-4)...u have a positive slope of 1/3

thats gonna be 2nd answer choice

notka56 [123]3 years ago

5 0

Answer:

$y\leq \frac{1}{3}x-4$

Step-by-step explanation:

we know that

The solution of the inequality is the shaded area below the solid line

The slope of the line is positive

The y-intercept of the solid line is equal to $-4$

therefore

The inequality must be

$y\leq \frac{1}{3}x-4$

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A supplier of heavy construction equipment has found that new customers are normally obtained through customer requests for a sa

ziro4ka [17]

Answer:

The probability that it will take fewer than six customer contacts to clear the inventory is 0.8%.

Step-by-step explanation:

We have a probability of making an individual sale of p=0.15.

We have 4 units, so the probability of clearing the inventory with n clients can be calculated as:

$P=\dbinom{n}{4}p^4q^{n-4}=\dbinom{n}{4}0.15^4\cdot 0.85^{n-4}$

As we see in the equation, n has to be equal or big than 4.

In this problem we have to calculate the probability that less than 6 clients are needed to sell the 4 units.

This probability can be calculated adding the probability from n=4 to n=6:

$P=\sum_{n=4}^6P(n)=\sum_{n=4}^6 \dbinom{n}{4}0.15^4^\cdot 0.85^{n-4}\\\\\\P=0.15^4(\dfrac{4!}{4!0!}\cdot 0.85^{4-4}+\dfrac{5!}{4!1!}\cdot0.85^{5-4}+\dfrac{6!}{4!2!}0.85^{6-4})\\\\\\P=0.15^4(1\cdot0.85^0+5\cdot0.85^1+15\cdot0.85^2)\\\\\\P=0.00051(1+4.25+10.84)\\\\\\P=0.00051\cdot16.09\\\\\\P=0.008$