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3 years ago

Solve and simplify, typing the final answer as a fraction 2/5 × 6/7

Mathematics

1 answer:

MArishka [77]3 years ago

7 0

Answer:

The answer is $\frac{12}{35}$ .

Step-by-step explanation:

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What is the value of the angle marked with x?

LuckyWell [14K]

Answer:

116 degrees

Step-by-step explanation:

It is equal to the measure of the angle opposite of x.

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3 years ago

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1. What concepts of functions can you associate with the pictures?

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Step-by-step explanation:

what concepts of functions can you associate with the picture

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3 years ago

Justin cuts a 10-inch-long roll of cookie dough into 6 equal-length pieces. How many inches long is each piece of cookie dough?

mixas84 [53]

Answer: Each piece is 1 2/3 inch length.

Step-by-step explanation:

Total length of cookie dough =10-inch

Since it is being into divided into 6 equal-length pieces

We have that each piece will have=1 0-inch / 6 = 5/3 = 1 2/3 inch length.

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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

4 years ago

An object is moving at a speed of 2 feet every 8.5 minutes. Express this speed in meters per hour

jekas [21]

Answer:

4.30m/hr

Step-by-step explanation:

2 feet/8.5 mins is 0.235294118 feet per minute, which is 4.3030553875397 metres per hour, which to 2 s.f. is 4.30

8 0

3 years ago