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3 years ago

A b c or d? please i need help

Mathematics

2 answers:

butalik [34]3 years ago

8 0

Answer:D

Step-by-step explanation:D

denpristay [2]3 years ago

3 0

Answer:

The answer is D.

Step-by-step explanation:

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Please help, this chapter was on derivatives...

Usimov [2.4K]

<h3>Answer: ds/dt = 11</h3>

================================================

Work Shown:

Before we can use derivatives, we need to find the value of s when (x,y) = (15,20)

s^2 = x^2+y^2

s^2 = 15^2+20^2

s^2 = 225+400

s^2 = 625

s = sqrt(625)

s = 25

-----------

Now we can apply the derivative to both sides to get the following. Don't forget to use the chain rule.

s^2 = x^2 + y^2

d/dt[s^2] = d/dt[x^2 + y^2]

d/dt[s^2] = d/dt[x^2] + d/dt[y^2]

2s*ds/dt = 2x*dx/dt + 2y*dy/dt

2(25)*ds/dt = 2(15)*5 + 2(20)*(10)

50*ds/dt = 150 + 400

50*ds/dt = 550

ds/dt = 550/50

ds/dt = 11

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Side note: The information t = 40 is never used. It's just extra info.

8 0

2 years ago

It cost $36 per person if 75 people rent the bus. How much will it cost per person if 100 people rent.

motikmotik

find the total cost of the bus by multiplying 75 people by $36:

75 x 36 = $2,700

Now divide the total cost of the bus by 100 people:

2700 / 100 = 27

It will cost $27 per person.

8 0

3 years ago

A. The area of a circle is 113.1 ft to the 2nd power. What would the diameter have to be.

irina [24]

What are the answers

3 0

2 years ago

????????????????????

Svet_ta [14]

#2- yes because both can be divided or multiplied for example 3 and 9, u can multiply 3x3 to get 9 since the dilation is 3 then yes it is similar triangles

4 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