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3 years ago

8

There are currently 75 million cars in a certain country, decreasing by 4.5% annually. how many years will it take for this coun

try to have 53 million cars?

1 answer:

pishuonlain [190]3 years ago

6 0

75 x 4.5%= 3.375
75 - 53 = 22
22 / 3.375 = 6.5
it will take 6.5 yrs for this country to have 53 million cars

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Suppose that 50% of the watches produced by a certain factory are defective. A store buys a box with 400 watches produced by thi

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Answer:

$(a)\ P(x \ge 215)$

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Step-by-step explanation:

Given

$p = 0.50$ ---- proportion of watches with defects

$n = 400$ --- Number of watches

Solving (a): Represent at least 215 of 400 are defective

In inequalities, at least means: $\ge$

So, the probability is represented as: $P(x \ge 215)$

Solving (b): Calculate $P(x \ge 215)$

Normal or Poisson: Normal distribution is characterized by 2 parameters $\mu$ and $\sigma$ .

These two parameters can be easily calculated from the given parameters in the question. So, we solve using normal distribution

Start by calculating the mean

$\mu =np$

$\mu = 0.50 * 400$

$\mu = 200$

Calculate standard deviation

$\sigma = \sqrt{\mu (1 - p)$

$\sigma = \sqrt{200 * (1 - 0.50)$

$\sigma = \sqrt{200 * 0.50$

$\sigma = \sqrt{100$

$\sigma = 10$

By continuity correction, we have:

$x \to x - 0.5$

$x \to 215 - 0.5$

$x \to 214.5$

So, we have:

$P(x \ge 215) = P(x \ge 214.5)$

Calculating $P(x \ge 214.5)$ , we have:

$P(x \ge 214.5) = 1 - P(x < z)$

Calculate z score

$z = \frac{x - \mu}{\sigma}$

$z = \frac{214.5 - 200}{10}$

$z = \frac{14.5}{10}$

$z = 1.45$

So, we have:

$P(x \ge 214.5) = 1 - P(x < 1.45)$

Using the z score probability table, we have:

$P(x < 1.45) = 0.92647$

So, we have:

$P(x \ge 214.5) = 1 - 0.92647$

$P(x \ge 214.5) = 0.07353$

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Answer: 2. c 3. d

Step-by-step explanation:

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Suppose a sample of 30 employees from the manufacturing industry showed a sample mean of $23.89 per hour. Assume a population st

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Answer:

The<em> p</em>-value of the test is 0.1212.

Step-by-step explanation:

A one sample <em>z</em>-test can be performed to determine whether the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.

The hypothesis is defined as:

<em>H₀</em>: The mean hourly wage is same as the reported mean of $24.57 for the goods-producing industries, i.e. <em>μ</em> = $24.57.

<em>Hₐ</em>: The mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries, i.e. <em>μ</em> ≠ $24.57.

The information provided is:

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Compute the test statistic as follows:

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{23.89-24.57}{2.40/\sqrt{30}}=-1.55$

The test statistic value is, <em>z</em> = -1.55.

Compute the <em>p</em>-value of the test as follows:

$p-value=2\times P(Z$

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Answer:

340 ways

Step-by-step explanation:

Given:

Total number of oranges = 60

Number of rotten oranges = 4

Number of oranges picked = 3

Now, number of good oranges = Total number - Rotten oranges

= 60 - 4 = 56

Now, we need to pick at least two rotten oranges.

So, the possible outcomes can be as follows:

2 rotten oranges + 1 good orange = 3 oranges
3 rotten oranges + 0 good orange = 3 oranges

Now, number of ways of picking 'r' distinct objects from a total of 'n' objects is given as:

$^nCr=\frac{n!}{r!(n-r)!}$

Now, picking 2 rotten oranges from a total of 4 rotten oranges is:

$^4C_2=\frac{4!}{2!2!}=\frac{4\times 3\times 2}{4}=6$

Similarly, picking 3 rotten oranges from a total of 4 rotten oranges is:

$^4C_3 =\frac{4!}{3!\times1!}=\frac{4\times 3!}{3!}=4$

Now, picking 1 good orange from a total of 56 good oranges is:

$^{56}C_1=56$

Picking 0 good oranges means picking no good oranges.

Therefore, the total number of ways of picking at least 2 rotten oranges is the sum of the above two possibilities and is given as:

At least 2 rotten out of 3 picked = (2 rotten and 1 good) or 3 rotten

= 6 × 56 + 4

= 336 + 4 = 340 ways

Therefore, there are 340 ways of picking at least 2 rotten oranges when 3 oranges are picked from a total of 60 oranges.

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