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Valentin [98]
2 years ago
7

Order the numbers from least to greatest A.1.81. B.2. C.1.511 D.1.799. ​

Mathematics
2 answers:
Studentka2010 [4]2 years ago
8 0
From least to greatest it is C. 1.511 , D. 1.799, A. 1.81 and lastly B.2.
So C,D,A,B
zimovet [89]2 years ago
3 0
B, A, D, C
or 2, 1.81, 1.799, 1.511
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The dot plots below show the number of blooms on miniature rose bush plants at two garden supply stores.
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Answer:

C

Step-by-step explanation

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3 years ago
Consider the equation 9x+3=x−1. Squaring the left side and simplifying results in​ _______. Squaring the right side and simplify
bixtya [17]

Answer:

The first blank is 81x^{2}+9=x-1

The second blank is 9x+3=x^2-1

Step-by-step explanation:

For the first part, you must set up the equation (9x+3)^2       then you must multiply the exponent to the inside exponents 9^{2} x^{2}+3^{2} next, simplify.

81x^{2}+9=x-1

For part two, set up equation. x^{2} -1^{2} Then simplify 9x+3= x^{2} -1

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3 years ago
You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob
koban [17]

Answer:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) P=0.42

c) P=0.08

d)

X    |    P(X)

------------------

0    |    0.42

1     |    0.50

2    |    0.08

e) E(x)=0.66

s.d.=0.62

Step-by-step explanation:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) The probability of losing the first game is

P(G_1=L)=1-P(G_1=W)=1-0.4=0.6

The probability of losing the second game, given that the first game was lost is:

P(G_2=L|G_1=L)=1-P(G_2=W|G_1=L)=1-0.3=0.7

So the probability of losing both games is:

P(G_2=L\&G_1=L)=P(G_1=L)*P(G_2=L|G_1=L)=0.6*0.7=0.42

c) The probability of winning both games is:

P(G_2=W\&G_1=W)=P(G_1=W)*P(G_2=W|G_1=W)=0.4*0.2=0.08

d) The variable X can take values 0, 1 and 2.

X=0 is when both games are lost. This happens with probability P=0.42.

X=2 is when both games are won. This happens with probability P=0.08.

X=1 is when one game is won and the other is lost. This happens with probability P=1-0.42-0.08=0.50.

Then the table of probabilities become:

X    |    P(X)

------------------

0    |    0.42

1     |    0.50

2    |    0.08

e) The expected value is:

E(x)=\sum p_ix_i=0.42*0+0.50*1+0.08*2=0.00+0.50+0.16=0.66

The variance and standard deviation of x are:

V(x)=\sum p_i(x_i-E(x))^2\\\\V(x)=0.42(0-0.66)^2+0.50(1-0.66)^2+0.08(2-0.66)^2\\\\V(x)=0.42*0.4356+0.50*0.1156+0.08*1.7956\\\\V(x)=0.183+0.058+0.144=0.385\\\\\\\sigma=\sqrt{V(x)}=\sqrt{0.385}=0.62

The standard deviation can

5 0
3 years ago
Which of the following best describes an unpredictable event?​
Anna35 [415]

Answer:

B. The weather on a particular day a year from now

Step-by-step explanation:

We can only predict the weather in the near future, not long term or all time. The rest of the answer choices are predictable. We will always know the age of a person 10 years from now, we can predict the rating of the movie if we preview and watch it, and if a student studies enough/not enough we can predict the type of grade they will get on a test.

6 0
2 years ago
Read 2 more answers
Find the inverse of the function. f(x) = 6x3 - 8
Papessa [141]
\bf f(x)=y=6x^3-8\qquad inverse\implies 
\begin{array}{lll}
x=&6y^3-8\\
&\uparrow\\
&\textit{switched variables} 
\end{array}
\\\\\\
x+8=6y^3\implies \cfrac{x+8}{6}=y^3\impliedby taking\ \sqrt[3]{\qquad }
\\\\\\
\sqrt[3]{\cfrac{x+8}{6}}=\sqrt[3]{y^3}\implies \sqrt[3]{\cfrac{x+8}{6}}=y\impliedby f^{-1}(x)
7 0
2 years ago
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