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3 years ago

Help me please!!!! I will mark Brainliest

Mathematics

1 answer:

ryzh [129]3 years ago

8 0

Answer:

32 pi

Step-by-step explanation:

im sorry if im wrong im not the smartest haha

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In problems 6 and 7, write an inequality that describes each

Mazyrski [523]

6.
c is less than or equal to 1 because a dollar is the maximum value, therefore everything else in the store has to be less.
7.
If the population is at least a billion, it can only go up from there, therefore it is 1,000,000,000 is less than or equal to p

7 0

2 years ago

Read 2 more answers

Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste

lidiya [134]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$ In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$ And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

For this case we have the following distribution given:

X 3 4 5 6

P(X) 0.07 0.4 0.25 0.28

We can calculate the mean with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$

In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$

And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

3 0

3 years ago

Police estimate that 25% of drivers drive without their seat belts. If they stop 6 drivers at random, find the probability tha

Furkat [3]

Answer:

17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$