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1 year ago

A Store-It-All storage facility requires users to enter a four digit code, using numbers 0-9, to get through the

Mathematics

2 answers:

den301095 [7]1 year ago

6 0

Answer:

Step-by-step explanation: 1. 5040 because 10x9x8x7

3. 10000 because 10^4=1000

sveta [45]1 year ago

3 0

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enyata [817]

Answer:

The value of r to be out in the given formula is r = 0.048.

Step-by-step explanation:

The rate is given as 4.8%.

Now to put the value of interest in the given formula,

We need the value of r.

r = $\frac{r \%}{100} = \frac{4.8}{100} = 0.048$

Hence, the value of r to be out in the given formula is r = 0.048.

3 0

3 years ago

How do you figure out what is 1/4 of 10.

frosja888 [35]

You can multiply 10 by 1/4 or .25

10x.25=2.5

3 0

2 years ago

Read 2 more answers

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

erica [24]

Answer:

(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

Let $E_1, E_2,$ and $E_3$ be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

where $P(E_2/E_1)$ denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

$P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)$

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

$P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)$