Use scientific notation to rewrite the number:

wo balls are chosen randomly from an um containing 8 white, 4 black,and 2 orange balls. Suppose that we win $2 for each black ba

umka21 [38]

Answer:

The probability distribution is shown below.

Step-by-step explanation:

The urn consists of 8 white (W), 4 black (B) and 2 orange (O) balls.

The winning and losing criteria are:

Win $2 for each black ball selected.
Lose $1 for each white ball selected.

There are 8 + 4 + 2 = 14 balls in the urn.

The number of ways to select two balls is, ${14\choose 2}=91$ ways.

The distribution of amount won or lost is as follows:

Outcomes: WW WO WB BB BO OO

X: -2 -1 1 4 2 0

Compute the probability of selecting 2 white balls as follows:

The number of ways to select 2 white balls is, ${8\choose 2}=28$ ways.

The probability of WW is,

$P(WW)=\frac{n(WW)}{N}=\frac{28}{91}=0.3077$

Compute the probability of selecting 1 white ball and 1 orange ball as follows:

The number of ways to select 1 white ball and 1 orange ball is, ${8\choose 1}\times {2\choose 1}=16$ ways.

The probability of WO is,

$P(WO)=\frac{n(WO)}{N}=\frac{16}{91}=0.1758$

Compute the probability of selecting 1 white ball and 1 black ball as follows:

The number of ways to select 1 white ball and 1 black ball is, ${8\choose 1}\times {4\choose 1}=32$ ways.

The probability of WB is,

$P(WB)=\frac{n(WB)}{N}=\frac{32}{91}=0.3516$

Compute the probability of selecting 2 black balls as follows:

The number of ways to select 2 black balls is, ${4\choose 2}=6$ ways.

The probability of BB is,

$P(BB)=\frac{n(BB)}{N}=\frac{6}{91}=0.0659$

Compute the probability of selecting 1 black ball and 1 orange ball as follows:

The number of ways to select 1 black ball and 1 orange ball is, ${4\choose 1}\times {2\choose 1}=8$ ways.

The probability of BO is,

$P(BO)=\frac{n(BO)}{N}=\frac{8}{91}=0.0879$

Compute the probability of selecting 2 orange balls as follows:

The number of ways to select 2 orange balls is, ${2\choose 2}=1$ ways.

The probability of OO is,

$P(OO)=\frac{n(OO)}{N}=\frac{1}{91}=0.0110$

The probability distribution of X is:

Outcomes: WW WO WB BB BO OO

X: -2 -1 1 4 2 0

P (X): 0.3077 0.1758 0.3516 0.0659 0.0879 0.0110

3 0

4 years ago

Bonjour est que quelqu'un pourrait m'aider svp pour cette exercice des mathématique Merci

Ivanshal [37]

(Le français n'est pas ma langue maternelle, alors je m'excuse d'avance si quelque chose n'est pas clair!)

Il me semble que le triangle MNP devrait ressembler à celui que j’ai attaché.

Les triangles MNP et MQR sont similaires, les côtés correspondants sont donc proportionnels:

$\dfrac{MQ}{MN}=\dfrac{MR}{MP}=\dfrac{QR}{NP}$

$\implies\dfrac{2,5}6=\dfrac{MR}{5,7}=\dfrac{1,9}{NP}$

$\dfrac{2,5}6=\dfrac{MR}{5,7}\implies MR=2,375$

$\dfrac{2,5}6=\dfrac{1,9}{NP}\implies NP=4,56$

3 0

4 years ago

Please Help!! Which graph shows the solution to this system of inequalities?

scoundrel [369]

Answer:

A. The first graph

Step-by-step explanation:

When you graph these two equations on a graphing calc, you should be able to see one dotted line and one solid, as well as the solutions being mostly in the 1st quadrant.

7 0

4 years ago

Which expression is equivalent to 4√x^10 x^2(4√x^2) x^2.2 x^3(4√x)

kotykmax [81]

$\sqrt[n]{a^m}=\a^\frac{m}{n}$

$\sqrt[4]{x^{10}}=x^\frac{10}{4}=x^{2\frac{2}{4}}=x^2\cdot x^\frac{2}{4}=x^2\sqrt[4]{x^2}$

$Used:\\\\a^n\cdot a^m=a^{n+m}$

4 0

3 years ago

HELP!! Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Rashid [163]

Answer:

20

85%

Step-by-step explanation:

You are given the function $S(n)=20\cdot b^n.$

If n is the number of hours, then initially n=0 and

$S(0)=20\cdot b^0=20\cdot 1=20.$

If S(n) is the function of exponential growth, then it can be represented as

$S(n)=I\cdot (1+r)^n,$

where I is the initial amount, r -is the percent growth rate and n is the number of hours.

If b = 1.85, we can represent it as b = 1 + 0.85. Thus, the hourly percent growth rate of the bacteria would be 0.85=85%.

8 0

3 years ago

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