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nalin [4]
3 years ago
12

Patricia is writing a book. If she writes an

Mathematics
1 answer:
kiruha [24]3 years ago
3 0
0.036 that would be the answer
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leila saw that the dress was marked down 30 percent during a sale. if the original price was $130 what was the sale price of the
pychu [463]

Answer:

The sale price of the dress is $91

Step-by-step explanation:

leila saw that the dress was marked down 30% during a sale. If the original price was $130.

Which means,

130 × 30 ÷ 100 = $39

So,

$130 – $39 = $91

Thus, The sale price of the dress is $91

<u>-TheUnknownScientist 72</u>

6 0
3 years ago
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Consider the polynomial equation x(x-3)(x+6)(x-7)=0 which of following are zeros in the equation?
Sergio [31]

Answer:

x=0,3,-6,7

Step-by-step explanation:

set each factor equal to zero and then solve for x.

3 0
4 years ago
What is the answer to -5(-3w-8)=21 for w
Sauron [17]

Answer:

w = -1.266

Step-by-step explanation:

distribute the 5 by everything in the parenthesis and then bring everyhting down then you should have a regular equation to work w

8 0
4 years ago
50 is decreased to 19. what is the percent of change?
Delicious77 [7]
62% percent change is it
8 0
3 years ago
A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa
UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so \mu = \frac{14}{60} = 0.2333

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0
4 years ago
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