What are the answers to number 4 and 5??

Assume that random guesses are made for seven multiple choice questions on an SAT test, so that there are nequals7 trials, eac

Ket [755]

Answer:

$P(X < 4)=0.1335+0.3115+0.3115+0.1730=0.9295$

Step-by-step explanation:

Question assumed: Find the probability that the number x of correct answers is fewer than 4.

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=7, p=0.25)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X < 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(7C0)(0.25)^0 (1-0.25)^{7-0}=0.1335$

$P(X=1)=(7C1)(0.25)^1 (1-0.25)^{7-1}=0.3115$

$P(X=2)=(7C2)(0.25)^2 (1-0.25)^{7-2}=0.3115$

$P(X=3)=(7C3)(0.25)^3 (1-0.25)^{7-3}=0.1730$

$P(X < 4)=0.1335+0.3115+0.3115+0.1730=0.9295$

3 0

4 years ago

Lucas is applying to both Stanford and Cal-poly. The probability that he gets accepted to Stanford is 0.4. The probability that

Bas_tet [7]

a) What is the probability that he is accepted to one of the two schools

this means P( standford or cal-poly)

P( standford or cal-poly) = P( standford) + P(cal-poly) − P( standford and cal-poly)

= 0.4 + 0.3 – 0.15

= 0.55

What is the probability that he does not get accepted at either school?

Prob that he gets neither = 1 − 0.55 = 0.45

7 0

3 years ago

How many sets of three integers between 1 and 20 are possible if no two consecutive integers are to be a set?

erastovalidia [21]

There are $\dbinom{20}3=1140$ total possible ways to pick any three integers from the set.

Of the total, there are 18 consisting of consecutive triplets $(\{1,2,3\},\{2,3,4\},\ldots,\{18,19,20\})$ .

Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs $(\{1,2\},\{2,3\},\ldots,\{19,20\})$ , and for each pair 18 possible choices for the third integer (for instance, $\{1,2\}$ can be taken with 3, 4, ..., 20), to a total of $19\times18=342$ . To avoid double-counting (e.g. $\{1,2\}$ can't go with 3; $\{2,3\}$ can't go with 1 or 4), we subtract 1 from the extreme pairs $\{1,2\}$ and $\{19,20\}$ (twice), and 2 from the rest (17 times).

So, the number of triplets that don't consist of pairwise consecutive integers is

$1140-(18+342-(2\times1+17\times2))=816$

I don't know how useful this would be to you, but I've verified the count in Mathematica:

In[8]:= DeleteCases[Subsets[Range[1, 20], {3}], x_ /; x[[2]] == x[[1]] + 1 || x[[3]] == x[[2]] + 1] // Length
Out[8]= 816

6 0

3 years ago

Solve the quadratic equation by factoring and show work: 6x²−x=15

alina1380 [7]

Answer:

$(2x + 3)(3x - 5)$