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3 years ago

Write the decimal expansion for seven ninths

Mathematics

2 answers:

Korolek [52]3 years ago

7 0

0.7777777 is the answer I got

AysviL [449]3 years ago

6 0

Answer:

On my calculator it appears as 0.7 recurring

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The table below represents a linear function f(x) and the equation represents a function g(x):

gulaghasi [49]

First, lets solve f(x). g(x) is already given. Find the slope of f(x) by doing m = y2-y1/x2-x1. I'm going to use -1,-5 (1) and 0,-1 (2). -1-(-5) = 4 and 0-(-1) = 1. This means that the slope is 4 since 4/1 = 4.
Now find y intercept (b). Since we know that y=4x+b, we can solve this by plugging in an ordered pair. I will use -1,-5. -5 = 4(-1) +b. With this, we see that -5 = -4+b and b = -1 since -5 = -4-1.
Your equation for f(x) is y=4x-1.
A: The slopes are equal since both of them are 4.
B: g(x) has a greater y intercept since 3 > -1.

3 0

3 years ago

Arianna went to the store to buy some chicken. The prince per pound of the chicken is $5.75 per pound and she has a coupon for $

gregori [183]

Answer:

$20

Step-by-step explanation:

If you buy 4 pounds of chicken, you multiply 5.75 times 4. That gives you $23. You have a coupon for $3 off the final amount. $23-$3 = $20.

6 0

3 years ago

Read 2 more answers

E MULTIPLE CHOICE QUESTION Now what does it look like after the next step? (hint: PEMDAS)

Sedbober [7]

Answer: Multiplication

Step-by-step explanation: The M in PEMDAS stands for Multiplication

8 0

3 years ago

What number should go in box A on this number line? 4. 5. A

cestrela7 [59]

Answer:

Step-by-step explanation:

4 0

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

3 years ago