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3 years ago

A company has 200 machines. Each machine has 12% probability of not working.

Mathematics

1 answer:

Eduardwww [97]3 years ago

4 0

1) This probability is given by

... C(40,5)·0.12⁵·0.88³⁵ ≈ 0.18665 . . . . . where C(n, k) = n!/(k!(n-k)!)

2) This probability is given by

... 0.88⁴⁰ ≈ 0.00602

3) This is the complement of the probability that all have failed.

... 1 - 0.12⁴⁰ ≈ 1.0000

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What is the value of f(3)?

Anni [7]

Answer:

-12

Step-by-step explanation:

Note that when x < 6, you will use the expression -4x

f(x) = -4x

Plug in 3 for x

f(3) = -4(3)

Simplify. Multiply

f(3) = -4(3)

f(3) = -12

-12 is your answer

5 0

3 years ago

Assume that thermometer readings are normally distributed with a mean of degrees and a standard deviation of 1.00degrees C. A th

Andrew [12]

Answer:

0.2684 is the probability that the temperature reading is between 0.50 and 1.75.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0 degrees

Standard Deviation, σ = 1 degrees

We are given that the distribution of thermometer readings is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(Between 0.50 degrees and 1.75 degrees)

$P(0.50 \leq x \leq 1.75)\\\\ = P(\displaystyle\frac{0.50 - 0}{1} \leq z \leq \displaystyle\frac{1.75-0}{1})\\\\ = P(0.50 \leq z \leq 1.75)\\= P(z \leq 1.75) - P(z < 0.50)\\= 0.9599 - 0.6915 = 0.2684 = 26.84\%$

0.2684 is the probability that the temperature reading is between 0.50 and 1.75.

7 0

3 years ago

If a^ 1/n = n What is an equivalent form of 7? Show steps plz!

Scrat [10]

Answer:

7⁷

Step-by-step explanation:

a^(1/7) = 7

a = 7⁷

6 0

3 years ago

Expand the logarithm as much as possible using properties. please help!

Solnce55 [7]

Step-by-step explanation:

log (√1000000x)

Rewrite √1000000x as (1000000x)1/2.

expand long ((1000000x)1/2) by moving 1/2

oby moving logarithm.

1/2 longth (1000000x)

Rewrite

log

(1000000x) as log(1000000)+log(x).

1/2(log(1000000)+log(x))

Logarithm base 10 of 1000000 is 6.

1/2(6+log(x))

Apply the distributive property.

1/2.6+1/2 log(x)

Cancel the common factor of 2.

3+1/2 long(x)

Combine 1/2 and log(x)

3+ long(x)/2

3 0

3 years ago

Read 2 more answers

Suppose the horses in a large stable have a mean weight of 1467lbs, and a standard deviation of 93lbs. What is the probability t

krok68 [10]

Answer:

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 1467, \sigma = 93, n = 49, s = \frac{93}{\sqrt{49}} = 13.2857$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable?

This is the pvalue of Z when X = 1467 + 9 = 1476 subtracted by the pvalue of Z when X = 1467 - 9 = 1458.

X = 1476

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1476 - 1467}{13.2857}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 1458

$Z = \frac{X - \mu}{s}$

$Z = \frac{1458 - 1467}{13.2857}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

5 0

3 years ago