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4 years ago

What is the term-to-term rule for this sequence: 64, 16, 4, 1, 0.25, …?

Mathematics

2 answers:

lina2011 [118]4 years ago

7 0

Answer: 0.0625

Step-by-step explanation:

0.25/4

Komok [63]4 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

Each term in the sequence is one- quarter of the previous term, thus

term to term rule is ' divide by 4 '

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Which of the following is a true statement about skew lines?

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A.) would be your answer

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3 years ago

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Please help! I'll rate 5 stars, if possible give brainliest, and give a thanks.

muminat

Answer:

2x+20 + x+40 = 180

Step-by-step explanation:

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2x+20 + x+40 = 180

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If x=9 and y=−3, evaluate the following expression:20−5y+2x

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Answer:

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3 years ago

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Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Please help i am very confused

Step2247 [10]

Answer:

$\sqrt{80}$

80 is the number in the square root

Step-by-step explanation:

$d = \sqrt{(x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} } \\d = \sqrt{(4-(-4) )^{2} +(5- 1 )^{2} }\\d = \sqrt{(4+4 )^{2} +4^{2}\\d = \sqrt{8^{2} +4^{2} }\\d=\sqrt{64 +16}\\d=\sqrt{80}$

4 0

3 years ago