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3 years ago

CAN SOMEONE PLEASE HELP ME WITH THIS LIKE PLEASE I NEED ANSWERS

Mathematics

1 answer:

EastWind [94]3 years ago

7 0

I’m sorry I can only answer number 3. The coordinates should be 1,-5. I am really not sure tho, because I have never done anything like this

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Can someone please help me please I really need help please help me.

kvv77 [185]

Answer:

Step-by-step explanation:

I can't get to act correctly either.

The answer is 1 day and 7 items.

Ariana

y = 5x + 2

Julie

y = 7x

The ys have to be the same so equate the right hand side.

7x = 5x + 2 Subtract 5x from both sides.

7x - 5x = 5x - 5x + 2

2x = 2 Divide by 2

2x/2 = 2/2

x = 1

1 day

7 items

6 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

Martin keeps track of the low temperature every day. On Monday, it was -12˚ On Tuesday, it was -9˚C. What is the difference betw

gogolik [260]

The difference is 3 declares Celsius.

6 0

3 years ago

Read 2 more answers

Use your creativity... Write a short story/problem/question that can be described by the scenario of 4 divided by 1/2

vazorg [7]

Answer:

I was at the store getting groceries I saw that there was a sell on bags of candy The original price was 4.00 dollars The sale is 1/2 off a bag of candy what is the new price. The answer is 2.00 dollars.

8 0

3 years ago

Choose as many answers as apply.

alexandr1967 [171]

Answer:

multiplying a multi-digit number by itself several times (finding the power of a number)
finding a square root
statistical calculations

Step-by-step explanation:

We don't know what your introduction tells you, but the above-listed operations are ones I choose to use a calculator for. I also use a calculator for ordinary arithmetic, such as division by numbers with 2 digits or more. (It is simply faster and requires no scratch paper.)

If statistical calculations are not done with a calculator, they at least require the availability of suitable tables.

___

All of these operations can be done by hand without a calculator, and were in times passed. Lifetimes of effort were involved in generating some of the original math tables for statistics, trig, logarithms, and other functions readily evaluated using a modern calculator.

3 0

3 years ago