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2 years ago

PLSS HELP ME ASAP!!! NO LINKS!!

Mathematics

1 answer:

kozerog [31]2 years ago

3 0

Answer:

0.25

because sum of probabilities is always equal to 1

i.e P(sean doesn't miss the bus)+P(sean misses the bus)=1

P(sean misses the bus) = 1-0.85 =0.25

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How many ways are there to distribute 5 identical apples and 6 identical pears to 3 distinct people such that each person has at

Nady [450]

Answer:

961

Step-by-step explanation:

5 apples and 6 pears are to be distributed among 3 different people.

the number of ways the fruits could be distributed without restrictions is 3^ 11 = 177147

If each person gets at least one pear,first calculate the number of ways 6 pears could be distributed among 3 people

= 3^6 - 3

= 729 -3 = 726

Secondly calculate the number of ways 5 apples could be distributed among 3 people

3^5 = 243

Total = 726 + 243

= 961 ways

6 0

3 years ago

20th term of 12, 15, 18, 21

Rainbow [258]

Answer:

Step-by-step explanation:

The rule: 3n+9

n=20

3(20)+9

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7 0

3 years ago

Evaluate the formula t=2n when n= 3.14, L=4, and g=44

dalvyx [7]

T = 2n, n = 3.14
t = 2(3.14)
t = 6.28

5 0

3 years ago

Work in teams of three or four (person A, person B, and person C). Each student is to complete his or her worksheet using the me

Marrrta [24]

Answer:

i need the answer as well

3 0

3 years ago

A product is made up of components A, B, C, D, E, F, G, H, I, and J. Components A, B, C, and F have a 1/10,000 chance of failure

lesantik [10]

Answer:

The overall reliability is 99.7402 %

Step-by-step explanation:

The overall reliability of the product is calculated as the product of the working probability of the components.

For components A,B,C and F we have :

$P(failure)=\frac{1}{10000}$

⇒

$P(Work)=1-\frac{1}{10000}=\frac{9999}{10000}=0.9999$

For components D,E,G and H we have :

$P(failure)=\frac{3}{10000}$

⇒

$P(Work)=1-\frac{3}{10000}=\frac{9997}{10000}=0.9997$

Finally, for components I and J :

$P(failure)=\frac{5}{10000}$

⇒

$P(Work)=1-\frac{5}{10000}=\frac{9995}{10000}=0.9995$

Now we multiply all the working probabilities. We mustn't forget that we have got ten components in this case :

Components A,B,C and F with a working probability of 0.9999

Components D,E,G and H with a working probability of 0.9997

Components I and J with a working probability of 0.9995

Overall reliability = $(0.9999)^{4}(0.9997)^{4}(0.9995)^{2}=0.997402$

0.997402 = 99.7402 %

4 0

3 years ago