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3 years ago

Find the mean median and mode(s) of the data shown:

Mathematics

1 answer:

AveGali [126]3 years ago

7 0

Median: 5 3/4 ( in a number line from least to greatest, the middle number)
Mode: 6 5/8 ( the number that appears the most [ex: 1,1,2,2,2,3,3 your mode will be 3])

hope this helps

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Need help with #21 please this is Algebra 2

kap26 [50]

You would do the amount x multiplier3
So it would be 5000 x (1.035)cubed
The answer is 5,543.59 I think

4 0

3 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

3 years ago

given the two points (-24,7) and (30,25) a. What is an equation passing through the points? b. Is (51, 33) also on the same line

kicyunya [14]

Part 1) Given the two points (-24,7) and (30,25) a. What is an equation passing through the points?

step 1
find the slope m
m=(y2-y1)/(x2-x1)---->m=(25-7)/(30+24)----> m=18/54----> m=1/3

step 2
wit m=1/3 and the point (30,25)
find the equation of the line
y-y1=m*(x-x1)-----> y-25=(1/3)*(x-30)--->y=(1/3)*x-10+25
y=(1/3)*x+15

the answer Part 1) is
y=(1/3)*x+15

Part 2) Is (51, 33) also on the same line?
if the point (51.33) is on the line y=(1/3)*x+15
then
for x=51 the value of y must be 33
for x=51
y=(1/3)*51+15----> y=17+15----> y=32
32 is not 33
so
the point does not belong to the given line

the answer Part 2) is
the point does not belong to the given line

see the attached figure

7 0

3 years ago

4x^2=27x+40 Solve by factoring

worty [1.4K]

Answer: hewo, there! your answer is below

x= -5/4

or x= 8

Step-by-step explanation:

step 1: Subtract 27x+40 from both sides.

4x2−(27x+40)=27x+40−(27x+40)

4x2−27x−40=0

Step 2: Factor left side of equation.

(4x+5)(x−8)=0

Step 3: Set factors equal to 0.

4x+5=0 or x−8=0

hope this helps you

have a great Day

Plz makr branilest

8 0

2 years ago

PLEASE HELP ASAP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! PLEASEEEEEEEEEE AND DO NOT LOOK IT UP ON ANY SOURCE THANK YOU AND ILY

AleksandrR [38]

I would go with Jesses survey.

Unlike everyone else’s survey Jesse chose 6th grade students in random, causing a more unbiased research.

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3 years ago

Read 2 more answers