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2 years ago

14

At an ice cream shop, 2 of the last 14 cones sold had cookie dough ice cream. what is the experimental probability that the next

cone sold will have cookie dough ice cream?

1 answer:

Kay [80]2 years ago

7 0

1 $\frac{1}{7}$

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∠DEF and ∠FEG are supplementary. m∠DEF=(3x+5)°, and m∠FEG=(2x)°. What is the measure of ∠DEF?

Alex787 [66]

Answer:

DEF = 110

Step-by-step explanation:

Supplementary angles add to 180

3x+5 +2x = 180

Combine like terms

5x +5 = 180

Subtract 5 from each side

5x = 175

Divide each side by 5

5x/5 = 175/5

x =35

DEF = 3x+5

= 3*35 +5

= 105+5

= 110

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2 years ago

Which value for x makes the sentence true?<br><br> 18x = 360

aniked [119]

Answer:

X=20

Step-by-step explanation:

360/18=20

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2 years ago

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I have a pet golden retriever. Each year it gains 11.4 pounds. He is 8 1\2 years old how many pounds does he weigh?

grigory [225]

I think he weighs 461.7 but I would check that

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2 years ago

remy said the table shows a proportional relationship because each distanc went up by 6ftg. Collin said that it was not a propor

sdas [7]

Answer:

remy would be correct

Step-by-step explanation:

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2 years ago

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

2 years ago