All categories

3 years ago

PLZ CAN SOMEONE HELP ME!!!!!!!

Mathematics

1 answer:

Hunter-Best [27]3 years ago

5 0

The answer is 4/16. The fastest way to solve these sorts of equations is to remember that a fraction is merely a division problem. Take the top number (numerator) and divide it by the bottom number ( denominator). That will give you a decimal. 4/16 is equal to .25. 5/20 is also equal to .25. You could also go about it by reducing the fraction. 5 goes into 20 4 times 5 goes into itself 1 time. so you have 1/4. 4 goes into 16 four times and into itself once. It is also 1/4. I hope this helps. :)

You might be interested in

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 4.

Leno4ka [110]

Answer:

P(4 non-work-related e-mails)=0.1933

P(7 or more non-work-related e-mails)=0.1442

P(3 or less non-work-related e-mails)=0.3771

Step-by-step explanation:

A Poisson distribution is useful to measure the probability of a number of events in a given period, in this problem, receiving a non-work-related e-mail would be the event and an hour would be the period of time.

Its probability mass function is $f(k; \lambda) = P(x=k) = \frac{\lambda^k e^{-\lambda}}{k!}$ , being k the number of events and λ how many sucesses (events) are per period of time (4.3 in this case)

First, we calculate what is the probability of receiving exactly 4 e-mails:

$f(k; \lambda) =P(k=4) =\frac{4.3^4 e^{-\4.3}}{4!}=0.1933$

As k is a discrete variable, to know the joint probability of different number of events we add the probability of each value.

The probability of receiving 7 or more is the sum of P(7), P(8), ..., P(∞):

$P(k\geq 7 ) =\sum_{i=7}^{\infty}\frac{4.3^i e^{-\4.3}}{i!}=0.1442$

Last, the probability of receiving 3 or less is the sum of P(0), P(1), P(2) and P(3):

$P(k\leq3 ) =\sum_{i=0}^{3}\frac{4.3^i e^{-\4.3}}{i!}=0.3771$

3 0

3 years ago

What is the correct answer for the mathematics equation 9 + 10 = is that to simple or too hard to answer

Rus_ich [418]

hmmm it’s a pretty difficult question I think it might be 19 or maybe 21 I don’t know

6 0

2 years ago

Read 3 more answers

Someone help me, like I need it desperately

Ierofanga [76]

Answer:

S and U are centres of the respective circles.

a. Three radii of Circle S : SU , SR , ST

(Lines from the centre of the circle to its circumference)

b.Three radii of Circle U : US , UR , UT

(Lines from the centre of the circle to its circumference)

c.The radii of the two circles are equal to each other as they are congruent to each other.

3 0

3 years ago

carmen bought 15 case of juice for a party. she bought 3 cases of bottled water for every 5 case of juice she bought for the par

Crazy boy [7]

Nine I believe, because fifth teen divided by five equals three
three times three equals nine.

3 0

3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago