Simplify: (-5a³-7a² +3)+(-9a³ +5a² +9)

Mathematics

1 answer:

DedPeter [7]2 years ago

3 0

Answer: =−14a3−2a2+12

Step-by-step explanation:

−5a3−7a2+3−9a3+5a2+9

=−5a3+−7a2+3+−9a3+5a2+9

Combine Like Terms:

=−5a3+−7a2+3+−9a3+5a2+9

=(−5a3+−9a3)+(−7a2+5a2)+(3+9)

=−14a3+−2a2+12

Answer:

=−14a3−2a2+12

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The joint probability mass function of XX and YY is given by p(1,1)=0p(2,1)=0.1p(3,1)=0.05p(1,2)=0.05p(2,2)=0.3p(3,2)=0.1p(1,3)=

inessss [21]

Answer:

P(Y=1|X=3)=0.125

Step-by-step explanation:

Given :

p(1,1)=0

p(2,1)=0.1

p(3,1)=0.05

p(1,2)=0.05

p(2,2)=0.3

p(3,2)=0.1

p(1,3)=0.05

p(2,3)=0.1

p(3,3)=0.25

Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)

P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)

P(X=3)=p(3,1) +p(3,2) +p(3,3)

P(X=3)=0.05+0.1+0.25=0.4

$P(Y=1|X=3)=\frac{P(X=3,Y=1)}{P(X=3)} =\frac{p(3,1)}{P(X=3)}=\frac{0.05}{0.4}= 0.125$

Hence P(Y=1|X=3)=0.125

8 0

3 years ago

Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an lg sm

GuDViN [60]

Answer: 0.951%

Explanation:

Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.

Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by

$P(x = k) = \frac{n!}{k!(n-k)!}p^k (1-p)^{n-k}$

Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones).

Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by

$P(x \ \textless \ 3) = P(x = 0) + P(x = 1) + P(x = 2)
\\ \indent = \frac{12!}{0!(12-0)!}(0.54)^0 (1-0.54)^{12-0} + \frac{12!}{1!(12-1)!}(0.54)^1 (1-0.54)^{12-1} + \\ \indent \frac{12!}{2!(12-2)!}(0.54)^2 (1-0.54)^{12-2}
\\
\\ \indent = \frac{12!}{(1)(12!)}(0.46)^{12} + \frac{12(11!)}{(1)(11!)}(0.54)(0.46)^{11}+ \frac{12(11)(10!)}{(2)(10!)}(0.54)^2(0.46)^{10}
\\
\\ \indent = (1)(0.46)^{12} + (12)(0.54)(0.46)^{11}+ (66)(0.54)^2(0.46)^{10}
\\ \indent \boxed{P(x \ \textless \ 3) \approx 0.00951836732 }
$

Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.

5 0

3 years ago

Condense the expression to the logarithm of a single quantity.<br><br>

vladimir1956 [14]

Answer:

$\log _3\left(\frac{x^{\frac{1}{2}}}{\left(y+8\right)^2}\right)$