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Romashka-Z-Leto [24]
3 years ago
6

The repairs bill on your car includes $355 for parts $146 for labor and a sales tax of $40 what is the total amount you owed

Mathematics
1 answer:
marissa [1.9K]3 years ago
7 0

Answer:

$541

Step-by-step explanation:

Given:

The repair bill includes $355 for parts $146 for labor and a sales tax of $40

Question asked :

what is the total amount we owed = ?

Solution:

Total cost of parts to repair = $355

Total cost of labor = $146

Sales tax =$40

By adding all the cost:

Total cost of repair = $355 + $146 + $40

                                = $541

Therefore, total amount we have to pay = $541

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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
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Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

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Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

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(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

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(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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