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Strike441 [17]
2 years ago
5

Christmas Tree lights are used not only for Christmas trees, but sometimes to decorate college dorm rooms. Suppose that a string

has 20 light bulbs. These lights are all wired in series, so that if one light bulb fails, the whole string fails. The probability that any one light fails is 0.02. Each of the individual lights are independent of each other. What is the probability that the string will NOT fail?
a. 0.02
b. 0.40
c. 0.668
d. 0.333
Mathematics
1 answer:
puteri [66]2 years ago
6 0

Answer: Hello!

Ok, because the bulbs are wired in series, then if only one fails, all the string fails.

Then we need to see the probability for the 20 bulbs to not fail.

If the probability for each bulb to fail is 0.02, then the complement (or the probability of working fine) is 1 - 0.02 = 0.98

then we have 20 bulbs, and each one has a probability of 0.98 of working alright, then the probability for all them to work alright is the multiplication of this probabilities, this is

0.98^{20} = 0.6676

rounded up in the decimal, we have 0.668

then the correct answer is c.

You might be interested in
Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $
nlexa [21]

Answer:

<em>$4.84</em>

<em></em>

Step-by-step explanation:

Given that

For delivery within 1\frac{1}2 mi, charges = $1.25

For delivery within 1\frac{1}2 mi - 1\frac{3}4 mi, charges = $1.70

For delivery within 1\frac{3}4 mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2\frac{1}4 mi, charges = $2.60

For delivery within 2\frac{1}4 mi - 2\frac{1}2 mi, charges = $3.05

For delivery within 2\frac{1}2 mi - 2\frac{3}4 mi, charges = $3.50

For delivery within 2\frac{3}4 mi - 3 mi, charges = $3.95

For delivery within 3 mi - 3\frac{1}4 mi, charges = $4.40

So, every 0.25 mi or \frac{1}4 mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of 3\frac{1}8 miles lies within the range 3 mi - 3\frac{1}4 mi.

So actual charges after increase of 10%

\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}

6 0
2 years ago
Determine whether the dilation from the figure on the left to the figure on the right is an enlargement or a reduction. Then fin
dybincka [34]

Answer:

A.

Step-by-step explanation:

we are going from a large shape to a smaller shape. the proof is that the side lengths go down from 7m to 0.5m. so, it is a reduction.

and the scaling factor s ? what do we need to multiply the original side lengths with to get the new ones ?

7 × s = 0.5 = 1/2

s = (1/2)/7 = (1/2)/(7/1) = (1×1)/(2×7) = 1/14

=> answer A is correct

8 0
2 years ago
A campus deli serves 300 customers over its busy lunch period from 11:30 a.m. to 1:30 p.m. A quick count of the number of custom
Irina-Kira [14]

Complete question is;

A campus deli serves 300 customers over its busy lunch period from 11:30 am to 1:30 pm. A quick count of the number of customers waiting in line and being served by the sandwich makers shows that an average of 10 customers are in process at any point in time. What is the average amount of time that a customer spends in process?

Answer:

4 minutes

Step-by-step explanation:

For this question, we will apply Little's law which is is a theorem that determines the average number of items in a stationary queuing system, based on the average waiting time of an item within a system and the average number of items arriving at the system per unit of time.

The formula for the law is:

Inventory = flow rate × flow time.

We are given;

Inventory = 300 customers

flow time is from 11: 30am to 1:30pm which is 2 hours = 120 minutes

Flow rate = 300/120 = 2.5 persons/minute

Now, Making flow rate the subject of the formula earlier given, we have;

flow rate =  inventory/ flow time

Flow time is the time each person spends in the process

Thus, plugging in the relevant values, we get ;

We are told that an average of 10 customers are in process at any point in time.

Thus;

Average flow time = average inventory/flow rate = 10/2.5 = 4 minutes

3 0
2 years ago
If f(x) = 5x + 40, what is f(x) when x = –5?<br><br> a.–9<br> b.–8<br> c.7<br> d.15
defon

Just plug the number

_5(5)+40=-25+40

=15

3 0
2 years ago
Read 2 more answers
A painter is painting a wall with an area of 150 ft2. He decides to paint half of the wall and then take a break. After his brea
Novosadov [1.4K]
The answer is 141.35 ft²

Before the first break, it was painted:
150 ft² ÷ 2 = 75 ft²
Now it's left:
150 ft² - 75 ft² = 75 <span>ft²

Before the second break, it was painted:
75 </span>ft² ÷ 2 = 37.5 <span>ft²
Now it's left:
75 </span>ft² - 37.5 ft² = 37.5 <span>ft²

Before the third break, it was painted:
37.5 </span>ft² ÷ 2 = 18.75 <span>ft²
</span><span>Now it's left:
</span>37.5 ft² - 18.75 ft² = 18.75 <span>ft²
</span>
<span>Before the fourth break, it was painted:
</span>18.75 ft² ÷ 2 = 9.375 <span>ft²
</span><span>Now it's left:
</span>18.75 ft² - 9.375 ft² = 9.375 <span>ft²
</span>
<span>Before the fourth break, it was painted:
</span>9.375 ft² ÷ 2 = 4.6875 <span>ft²
</span><span>Now it's left:
</span>9.375 ft² - 4.6875 ft² = 4.6875 ft²

Now, we will sum what he painted for now:
75 ft² + 37.5 ft² + 18.75 ft² + 9.375 ft² 4.6875 ft² = 141.3125 ft² ≈ 141.35 ft²

When the painter takes his fifth break, there will be <span>141.35 ft² of the wall painted.</span>
6 0
2 years ago
Read 2 more answers
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