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Sauron [17]
3 years ago
11

What is 4/8 simplified ​

Mathematics
1 answer:
Vinvika [58]3 years ago
8 0

Answer:

exact form: 1/2

decimal form: 0.5

Step-by-step explanation:

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What is the value of one striped present and one polka dot present
GarryVolchara [31]
Is there more to this question? If so, please revise your question so all your smart Brainly friends can help you!
4 0
3 years ago
PLEASE HELP WILL MARK BRAINLEIST
horsena [70]

Answer:

The picture shows the answer on the graph

Step-by-step explanation:

-2=-3/4×(x-6)

Distribute -3/4 through the parenthesis

-2= -3/4x+9/2

Multiply both sides of the equation by 4

-8=-3x+18

Move the variable to the left side and change its sign

3x-8=18

Move constant to the right side and change its sign

3x=18+8

Add the numbers

3x=26

Divide both sides of the equation by 3

x=26/3

Alternative Form- x=8 2/3 or x=8.6

7 0
3 years ago
An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample
valentina_108 [34]

Answer:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

Replacing we got:

\bar X=58.875 represent the mean

s=5.083 represent the sample standard deviation for the sample  

n=8 sample size  

\mu_o =60 represent the value that we want to test

\alpha=0.1 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:  

Null hypothesis:\mu \geq 60  

Alternative hypothesis:\mu < 60  

The statistic would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info we got:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0
3 years ago
What equation represents a parabola with a focus of (0,4) and a directrix of y=2
scZoUnD [109]
We know that

Any point <span>(x,y)</span> on the parabola is equidistant from the focus and the directrix

Therefore,

focus (0,4) and directrix of y=2

<span>√[<span>(x−0)</span></span>²+(y−4)²]=y−(2)

<span>√[x</span>²+(y-4)²]=y-2

x²+(y-4)²=(y-2)²

x²+y²-8y+16=y²-4y+4

x²=4y-12-----> 4y=x²+12----->y= (x²/4)+3


the answer is

y= (x²/4)+3


7 0
3 years ago
A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is
koban [17]

The expected length of code for one encoded symbol is

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

where p_\alpha is the probability of picking the letter \alpha, and \ell_\alpha is the length of code needed to encode \alpha. p_\alpha is given to us, and we have

\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of

\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375

bits to the code per encoded letter. For a string of length n, we would then expect E[L]=1.375n.

By definition of variance, we have

\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4

so that the variance for the length such a string is

\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859

"squared" bits per encoded letter. For a string of length n, we would get \mathrm{Var}[L]=1.859n.

5 0
3 years ago
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