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katrin [286]
3 years ago
6

Please help me out here. I know the answer but I don't know what the radical is​

Mathematics
1 answer:
PolarNik [594]3 years ago
4 0

Answer:

9

Step-by-step explanation:

There is no radical it's just 9

You might be interested in
276x^2−136x+869+−5184x+6
iragen [17]

Answer:

276x^{2}  -5320x +875

Step-by-step explanation:

Collect like terms and calculate the sum

7 0
3 years ago
A research team at Cornell University conducted a study showing that approximately 10% of all businessmen who wear ties wear the
LenKa [72]

Answer:

a) The probability that at least 5 ties are too tight is P=0.0432.

b) The probability that at most 12 ties are too tight is P=1.

Step-by-step explanation:

In this problem, we could represent the proabilities of this events with the Binomial distirbution, with parameter p=0.1 and sample size n=20.

a) We can express the probability that at least 5 ties are too tight as:

P(x\geq5)=1-\sum\limits^4_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\geq5)=1-(0.1216+0.2702+0.2852+0.1901+0.0898)\\\\P(x\geq5)=1-0.9568=0.0432

The probability that at least 5 ties are too tight is P=0.0432.

a) We can express the probability that at most 12 ties are too tight as:

P(x\leq 12)=\sum\limits^{12}_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\leq 12)=0.1216+0.2702+0.2852+0.1901+0.0898+0.0319+0.0089+0.0020+0.0004+0.0001+0.0000+0.0000+0.0000\\\\P(x\leq 12)=1

The probability that at most 12 ties are too tight is P=1.

5 0
3 years ago
34​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0
3 years ago
Harper has $14 in her piggy bank. Her mom pays her $12 for chores. How much does Harper now have in the piggy bank?
vovikov84 [41]

Answer:

26

Step-by-step explanation:

14+12=26

1+1=2

4+2=6

7 0
3 years ago
Read 2 more answers
g If the economy improves, a certain stock stock will have a return of 23.4 percent. If the economy declines, the stock will hav
dusya [7]

Answer:

E(X) = 23.4* 0.67 -11.9*0.33= 11.759 \%

Now we can find the second central moment with this formula:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = (23.4)^2* 0.67 +(-11.9)^2*0.33= 413.5965

And the variance is given by:

Var(X) = E(X^2) - [E(X)]^2

And replacing we got:

Var(X) = 413.5965 -(11.759)^2 =275.5105

And finally the deviation would be:

Sd(X) = \sqrt{275.5105}= 16.599 \%

Step-by-step explanation:

We can define the random variable of interest X as the return from a stock and we know the following conditions:

X_1 = 23.4 , P(X_1) =0.67 represent the result if the economy improves

X_2 = -11.9 , P(X_1) =0.33 represent the result if we have a recession

We want to find the standard deviation for the returns on the stock. We need to begin finding the mean with this formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing the data given we got:

E(X) = 23.4* 0.67 -11.9*0.33= 11.759 \%

Now we can find the second central moment with this formula:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = (23.4)^2* 0.67 +(-11.9)^2*0.33= 413.5965

And the variance is given by:

Var(X) = E(X^2) - [E(X)]^2

And replacing we got:

Var(X) = 413.5965 -(11.759)^2 =275.5105

And finally the deviation would be:

Sd(X) = \sqrt{275.5105}= 16.599 \%

7 0
3 years ago
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