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2 years ago

Can someone help me please

Mathematics

1 answer:

madreJ [45]2 years ago

4 0

Answer:

2<x<8

Here is the answer to he above question you asked. (I have to type at least 20 words to post my answer...)

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Solve for n: y=6nx – 10<br><br><br>And how to write the final answer

USPshnik [31]

Answer:

y=6nx-10

y+10=6nx

y+10/6x=n

therefore, n=y+10/6x.

Step-by-step explanation:

3 0

2 years ago

Can’t figure this out, I keep getting it wrong

uysha [10]

Two units, count the squares. Random letters because minimum word count

6 0

2 years ago

Find the values of the 30th and 90th percentiles of the data. Please show your work.

MaRussiya [10]

Answer:

30th percentile:109

90th percentile: 188

Step-by-step explanation:

The given data set is:

129, 113, 200, 100, 105, 132, 100, 176, 146, 152

We arrange the data set in ascending order to obtain;

Array= $100,100,105,113,129,132,146,176,200$

The 30th percentile is located at

$(\frac{30}{100}\times 10 )}$ -th position, where n=10 is the number of items in the data set

This implies that the 30th percentile is at the 3rd position.

Since 3 is an integer, the 30th percentile is the average of the 3rd and 4th occurrence in the array;

This implies that the 30th percentile = $\frac{105+113}{2}=\frac{218}{2}=109$

The 90th percentile is located at

$(\frac{90}{100}\times 10 )}$ -th position, where n=10 is the number of items in the data set

This implies that the 90th percentile is at the 9th position.

Since 9 is an integer, the 90th percentile is the average of the 9th and 10th occurrence in the array;

This implies that the 90th percentile = $\frac{176+200}{2}=\frac{376}{2}=188$

6 0

3 years ago

The exponential distribution applies to lifetimes of a certain component. Its failure rate is unknown. Find the probability that

wlad13 [49]

Answer:

a) 0.0821

b) 0.0111

c) 0.0041

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \lambda e^{-\lambda x}$

In which $\lambda$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\lambda x}$

Either it lasts more 5 years or less, or it survives more than 5 years. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

In all 3 cases, we want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = 1 - e^{-5\lambda}$

(a) lambda=.5

$P(X \leq 5) = 1 - e^{-5*0.5} = 0.9179$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9179 = 0.0821$

(b) lambda=0.9

$P(X \leq 5) = 1 - e^{-5*0.9} = 0.9889$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9889 = 0.0111$

(c) lambda=1.1

$P(X \leq 5) = 1 - e^{-5*1.1} = 0.9959$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9959 = 0.0041$

7 0

3 years ago

If an event has a 55% chance of happening in one trial, how do I determine the chances of it happening more than once in 4 trial

aev [14]

The chances of it happening more than once in 4 trials is 13%

<h3>How to determine the number</h3>

From the information given, we have can deduce that;

Probability of 1 trial = 55%

= 55/ 100

Find the ratio

= 0. 55

We are to find the probability of it happening more than once in 4 different trials

If the probability of it happening in one trial is 555 which equals 0. 55

Then the probability of it happening in 1 in 4 trials is given as;

P(1/4 trials) = 1/ 4 × 55%

P(1/4 trials) = 1/ 4 × 0. 55

Put in decimal form

P(1/4 trials) = 0. 25 × 0. 55

P(1/4 trials) = 0. 138

But we have to know the percentage

= 0. 138 × 100

Multiply the values, we have

= 13. 8 %

Thus, the chances of it happening more than once in 4 trials is 13%

Learn more about probability here:

brainly.com/question/24756209

#SPJ1

6 0

1 year ago