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pochemuha
3 years ago
11

Suppose​ a, b, and c are all negative numbers. How do you find the distance between the points​ (a,b) and​ (a,c)?

Mathematics
1 answer:
Illusion [34]3 years ago
6 0
Add the (a,b) and then add (a,c). Find out the number between these two after being added
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Let a = -4.82 and b = 4.35. All of the following statements are true except
kolezko [41]

Answer:

the sum of -4.82 and 4.35 = -0.47

so a is correct

the product of -4.82 and 4.35 = -20.967

so b is incorrect

the quotient of -4.82 and -4.35 is 1.10804597701

the quotient of 4.82 and 4.35 is 1.10804597701

so c is correct

if we subtract 4.35 from -4.82 we get -9.17

so d is incorrect

Hope This Helps!!!

6 0
3 years ago
Hospitals typically require backup generators to provide electricity in the event of a power outage. Assume that emergency backu
JulsSmile [24]

Answer:

a) There is a 10.24% probability that both generators fail during a power outage.

b) There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

Step-by-step explanation:

For each emergency backup generator, there are only two possible outcomes. Either they work correctly, or they fail. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

Assume that emergency backup generators fail 32% of the times when they are needed. So they work correctly 100-32 = 68% of the time. So p = 0.68

There are two generators, so n = 2

a. Find the probability that both generators fail during a power outage

This is P(X = 0)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.68)^{0}.(0.32)^{2} = 0.1024

There is a 10.24% probability that both generators fail during a power outage.

b. Find the probability of having a working generator in the event of a power outage. Is that probability high enough for the hospital? Assume the hospital needs both generators to fail less than 1% of the time when needed.

Either there are no working generators, or there is at least one working generator. 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) = 1 - 0.1024 = 0.8976

There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

To be high enough for the hospital, this probability should be at least of 99%.

7 0
4 years ago
What is the volume of the cylinder below?
FrozenT [24]

Answer:b

Step-by-step explanation:9 multiply 8 multiply 6

3 0
3 years ago
Read 2 more answers
The college student girls $3,000 for nine months to pay a semester of school if the interest is $185.63 find the monthly payment
Mademuasel [1]
3000 for 9 months...plus the interest of 185.63

(3000 + 185.63) / 9 = 3185.63 / 9 = 353.96 per month (and thats rounded)
8 0
3 years ago
What is the greatest common factor of the monomial 10x²y^5 and 15xy³? Explain
Eddi Din [679]
5x²y³ is the GCF
5 is the GCF of 10 and 15, to find the GCF of the variables, take the lowest exponent power of each of the variables
4 0
3 years ago
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