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LenKa [72]
3 years ago
12

Help me out please!!!!

Mathematics
2 answers:
Mariana [72]3 years ago
8 0
The answer is C (sorry if it’s wrong )
KonstantinChe [14]3 years ago
7 0

Answer:

c)

Step-by-step explanation:

- \frac{7}{ - f} \: is \: equal \: to \:  \frac{7}{f}

so \: the \: answer \: is \: none \: of \: the \: above

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What is 2+2736=344×2484848÷484848=
hammer [34]
2+2736=2738*2484848/484848=14032.26129
8 0
3 years ago
Solve using the quadratic formula
ch4aika [34]

Answer:

Solve the equation for  s  by finding  a  ,  b  , and  c  of the quadratic then applying the quadratic formula.  

s = − 2

Double roots

8 0
3 years ago
The density of oxygen is 0.001429 grams per cubic centimeter. How is this number written in
Anika [276]

We have been given that the density of oxygen is 0.001429 grams per cubic centimeter. We are asked to write the given number in scientific notation.

To convert a number into scientific notation, we write the given number as product of two numbers. One factor is a number between 1 and 10, while the 2nd factor is power of 10.  

When a decimal is to the left of given number, then we need to move the decimal to right such that the number gets between 1 and 10. Then, we multiply the number by negative power of 10 that many times, we moved the decimal to right.    

We can see that decimal is to left of our number. To place decimal after 1, we have to multiply 10 to 3 times. To keep the value of number same we will multiply the same negative power (negative 3).

0.001429\times 10^3\times 10^{-3}

(0.001429\times 10^3)\times 10^{-3}

(1.429)\times 10^{-3}

1.429\times 10^{-3}

Therefore, the number 0.001429 will be 1.429\times 10^{-3} in scientific notation.

7 0
3 years ago
|(2+1/3) + 0.3| = 3.5
nignag [31]

Answer:

False

Step-by-step explanation:

l(2+1/3) + 0.3| = 3.5

|7/3+3/10|=7/2

|79/30|=7/2

158=210

false

3 0
2 years ago
Read 2 more answers
If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distributio
goldenfox [79]

Answer:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

Step-by-step explanation:

For this case we have that the sample size is n =6

The sample man is defined as :

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

And we want a normal distribution for the sample mean

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

So for this case we need to satisfy the following condition:

X_i \sim N(\mu , \sigma), i=1,2,...,n

Because if we find the parameters we got:

E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu

Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}

And the deviation would be:

Sd (\bar X) = \frac{\sigma}{\sqrt{n}}

And we satisfy the condition:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

3 0
3 years ago
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