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OleMash [197]
2 years ago
5

Year Speed 1 38 2 46 3 52 4 57 5 61 3) During which time period was the ANNUAL rate of increase of the speed the GREATEST? HELP

FAST
A) from year 1 to year 2
B) from year 1 to year 3
C) from year 1 to year 4
D) from year 1 to year 5

Mathematics
2 answers:
Maksim231197 [3]2 years ago
8 0

Answer: A) from year 1 to year 2

Step-by-step explanation: subtract 46-38=8, I also just took the K12 test and got it right.

MatroZZZ [7]2 years ago
4 0

Answer:

A) from year 1 to year 2

Step-by-step explanation:

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Alona [7]

Answer:

Step-by-step explanation:

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6 0
3 years ago
A boiler has five identical relief valves. The probability that any particular valve will open on demand is 0.93. Assume indepen
noname [10]

Answer:

There is a 99.99998% probability that at least one valve opens.

Step-by-step explanation:

For each valve there are only two possible outcomes. Either it opens on demand, or it does not. This means that 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 combinatios 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:

n = 5, p = 0.93

Calculate P(at least one valve opens).

This is P(X \geq 1)

Either no valves open, 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

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

So

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

P(X = 0) = C_{5,0}.(0.93)^{0}.(0.07)^{5} = 0.0000016807

Finally

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000016807 = 0.9999983193

There is a 99.99998% probability that at least one valve opens.

5 0
3 years ago
Read 2 more answers
For every 3 kiwis there are 4
MArishka [77]

Answer:

Ratio of 3/4 I think

Step-by-step explanation:

Because you said for every 3 kiwis there are 4 of something so you would take the number of kiwis over the other number.

4 0
3 years ago
Need help i have been stuck on this problem
liberstina [14]
The second one is correct
4 0
4 years ago
"A manufacturer of automobile batteries claims that the average length of life for its grade A battery is 55 months. Suppose the
STALIN [3.7K]

Answer:

\\ P(z>-2) = 0.97725 or P(x>49) is about 97.725% (or being less precise 97.5% using the <em>empirical rule</em>).

Step-by-step explanation:

We solve this question using the following information:

  1. We are dealing here with <em>normally distributed data</em>, that is "<em>the frequency distribution of the life length data is known to be mound-shaped</em>".
  2. The normal distribution is defined by two parameters: the population mean (\\ \mu) and the population standard deviation (\\ \sigma). In this case, we have that \\ \mu = 55 months, and \\ \sigma = 3 months.
  3. To find the probabilities, we have to use the <em>standard normal distribution</em>, which has \\ \mu = 0 and \\ \sigma = 1. The probabilities for this distribution are collected in the <em>standard normal table</em>, available in Statistics books or on the Internet. We can also use statistics programs to find these probabilities.
  4. For most cases, we need to use the <em>cumulative standard normal table, </em>and for this we have to previously "transform" a raw score (x) into a z-score using the next formula: \\ z = \frac{x - \mu}{\sigma} [1]. A z-score tells us the distance from the mean that a raw score is from it in <em>standard deviations units</em>. If this value is <em>negative</em>, the raw score is <em>below</em> the mean. Conversely, a <em>positive</em> value indicates that it is <em>above</em> the mean.
  5. The <em>cumulative standard normal table </em>is made for positive values of z. Since the normal distribution is <em>symmetrical</em> around the mean, we can find the negative values of z using this formula: \\ P(z [2].

Having all this information, we can solve the question.

<h3>The percentage of the manufacturer's grade A batteries that will last more than 49 months</h3>

<em>First Step: Use formula [1] to find the z-score of the raw score x = 49 months</em>.

\\ z = \frac{49 - 55}{3}

\\ z = \frac{-6}{3}

\\ z = -2

This means that the raw score is represented by a z-score of \\ z = -2, which tells us that it is<em> two standard deviations below</em> the population mean.

<em>Second Step: Consult this value in the cumulative standard normal table for z = 2 and apply the formula [2] to find the corresponding probability.</em>

For a z = 2, the probability is 0.97725.  

Then

\\ P(z

\\ P(z2)

\\ P(z2)

But we <em>are not asked</em> for P(z<-2) but for P(z>-2) = P(x>49). This probability is the <em>complement</em> of the previous result, that is

\\ P(z>-2) = 1 - P(z

\\ P(z>-2) = 1 - 0.02275

\\ P(z>-2) = 0.97725

That is, the "<em>percentage of the manufacturer's grade A batteries will last more than 49 months</em>" is

\\ P(z>-2) = 0.97725 or about 97.725%

A graph below shows this result.

Notice that if we had used the <em>68-95-99.7 rule</em> (also known as the <em>empirical rule</em>), that is, in a normal distribution, the interval between <em>one standard deviation below and above the mean</em> contains, approximately, 68% of the observations; the interval between <em>two standard deviations below and above the mean</em> contains, approximately, 95% of the observations; and the interval between <em>three standard deviations</em> below and above the mean contains, approximately, 99.7% of the observations, we could have concluded that 2.5 % of the manufacturer's grade A batteries will last <em>less</em> than 49 months, and, as a result, 1 - 0.025 = 0.975 or 97.5% will last more than 49 months.

We can conclude that with a less precise answer (but faster) because of the <em>symmetry of the normal distribution</em>, that is, 1 - 0.95 = 0.05. At both extremes we have 0.05/2 = 0.025 or 2.5% and we were asked for P(x>49) = P(z>-2) (see the graph below).

6 0
4 years ago
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