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2 years ago

If the probability of a machine producing a defective part is 0.05, what is the probability of

Mathematics

1 answer:

Musya8 [376]2 years ago

6 0

Answer:

0.1800 to 4 places of decimals.

Step-by-step explanation:

Using the Binomial formula

Probability = 10C5* (0.95)^95 * (0.05)^5

= 100! / 95!*5! * (0.95)^95 * (0.05)^5

= 0.1800178.

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3 years ago

Hugo lives in the northwest corner of Greene County, Lena lives 29 miles away in the southeast corner, and the office where they

Annette [7]

Using the relation between velocity, distance and time, it is found that the commute took 48 minutes.

<h3>What is the relation between velocity, distance and time?</h3>

Velocity is distance divided by time, that is:

$v = \frac{d}{t}$

For Lena, we have that d = 29, v = 36, hence the time in hours is given by:

$t = \frac{d}{v} = \frac{29}{36} = 0.8056$

In minutes, the time is given by:

tM = 0.8056 x 60 = 48.

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1 year ago

A company borrows $75,000 for 8 years at a simple interest rate of 7.5%. Find the interest paid on the loan and the total amount

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The interest is $58,760.84
The total is $133,760.84

4 0

2 years ago

John bought a used truck for $4,500. He made an agreement with the dealer to put $1,500 down and make payments of $350 for the n

musickatia [10]

Answer:

(D) 36%

Step-by-step explanation:

Actual cost of truck = $4500

Down payment = $1500

Money left to be paid = 4500-1500 = $3000

A = Monthly payments = 350

n = 10

Principal value (P) = 3000

Using formula,

A= $\frac{Pr(1+r)^{n}}{(1+r)^{n}-1}$

350 = $\frac{3000r(1+r)^{10}}{(1+r)^{10}-1}$

= 36% (approx)

Hence (D) is correct.

5 0

2 years ago

The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There

KIM [24]

Answer:

If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 15

Standard Deviation, σ = 1

Sample size = 4

Total lifetime of 4 batteries = 40 hours

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

Formula:

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

Standard error due to sampling:

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5$

We have to find the value of x such that the probability is 0.05

P(X > x) = 0.05

$P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05$

$= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05$

$=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83$

Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

8 0

3 years ago