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GalinKa [24]
2 years ago
12

If 70% of college students say they have a job what is the probability that 3 randomly selected college students say they have a

job?
Mathematics
1 answer:
dsp732 years ago
4 0

Answer: 0.343

Step-by-step explanation:

Probability of college students that say they have a job = 70%

Number of randomly selected college students = 3

The probability that 3 randomly selected college students say they have a job will be:

= (70/100)³

= 0.7 × 0.7 × 0.7

= 0.343

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trapecia [35]

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C. -4

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Question 3 of 10 * - 3 is a factor of P(x) =*? – 73? + 158-9 O A. True O B. False
Elina [12.6K]

Given,

The expression is,

P(x)=x^3-7x^2+15x-9

If x- 3 is the factor of the given polynomial function, then on substituting the value of x = 3 in the polynomial function, it become zero.

Thus subsituting the value of x in the polynomial,

\begin{gathered} P(3)=3^3-7\times3^2+15\times3-9 \\ =27-7\times9+45-9 \\ =27+45-63-9 \\ =72-72 \\ =0 \end{gathered}

Here, the obtained value of polynomial is 0.

Hence, x-3 is a factor of the given polynomial.

Thus, option A(true) is correct.

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10 months ago
An certain brand of upright freezer is available in three different rated capacities: 16 ft3, 18 ft3, and 20 ft3. Let X = the ra
ruslelena [56]

Answer:

E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6

E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6=349.2

V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24

The expected price paid by the next customer to buy a freezer is $466

Step-by-step explanation:

From the information given we know the probability mass function (pmf) of random variable X.

\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|

<em>Point a:</em>

  • The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is

E(X) = $\sum_{x\in D} x\cdot p(x)

Therefore

E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6

  • If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by <em>E[h(X)]</em> is computed by

E[h(X)] = $\sum_{D} h(x)\cdot p(x)

So h(X) = X^2 and

E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2

  • The variance of X, denoted by V(X), is

V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2

Therefore

V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24

<em>Point b:</em>

We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:

From the rules of expected value this proposition is true:

E(aX+b)=a\cdot E(x)+b

We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore

The expected price paid by the next customer is

60\cdot E(X)-650=60\cdot 18.6-650=466

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3 years ago
Solve for f:5/6F=5<br> Please help me<br> I will give brainliest
vredina [299]

Answer:

f = 6

Step-by-step explanation:

5/6F=5

Multiply each side by 6/5 to isolate f

6/5 * 5/6F=5*6/5

f = 6

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3 years ago
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