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2 years ago

12

Im going to give 15 points for this question please answer it

1 answer:

pochemuha2 years ago

6 0

I think it is either 4.41 or 1.84 Sq Ft

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Suppose a bag of marbles has 4 green, 2 red, 5 yellow, 1 brown, and 7 blue marbles. What is the probability of picking a red mar

swat32

Assuming that each marble can be picked with equal probability, we notice that there is a total of

$4+2+5+1+7 = 19$

marbles, of which 2 are red.

So, the probability of picking a red marble is

$\dfrac{2}{19}$

In fact, as in any other case of (finite) equidistribution, we used the formula

$P(\text{event}) = \dfrac{\text{number of favourable cases}}{\text{number of all possible cases}}$

4 0

2 years ago

Consider a uniform distribution from aequals4 to bequals29. (a) Find the probability that x lies between 7 and 27. (b) Find th

weeeeeb [17]

Answer:

a) 80% probability that x lies between 7 and 27.

b) 28% probability that x lies between 6 and 13.

c) 44% probability that x lies between 9 and 20.

d) 28% probability that x lies between 11 and 18.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value x between c and d, in which d is larger than c, is given by the following formula.

$P(c \leq x \leq d) = \frac{d - c}{b - a}$

Uniform distribution from a = 4 to b = 29

(a) Find the probability that x lies between 7 and 27.

So $c = 7, d = 27$

$P(7 \leq x \leq 27) = \frac{27 - 7}{29 - 4} = 0.8$

80% probability that x lies between 7 and 27.

(b) Find the probability that x lies between 6 and 13. 

So $c = 6, d = 13$

$P(6 \leq x \leq 13) = \frac{13 - 6}{29 - 4} = 0.28$

28% probability that x lies between 6 and 13.

(c) Find the probability that x lies between 9 and 20.

So $c = 9, d = 20$

$P(9 \leq x \leq 20) = \frac{20 - 9}{29 - 4} = 0.44$

44% probability that x lies between 9 and 20.

(d) Find the probability that x lies between 11 and 18.

So $c = 11, d = 18$

$P(11 \leq x \leq 18) = \frac{18 - 11}{29 - 4} = 0.28$

28% probability that x lies between 11 and 18.

3 0

2 years ago

Jacob has several babysitting jobs.on one job ,he earned $10 an hour plus an extra $20 since there were 3 children.On another jo

Umnica [9.8K]

Answer:

4 hours, $60

Step-by-step explanation:

6 0

2 years ago

The modified accelerated cost recovery system (MACRS):

ZanzabumX [31]

bahala kau ha ha. Matt wahak

3 0

2 years ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

2 years ago