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Nata [24]
3 years ago
6

What is

leqslant - 15" align="absmiddle" class="latex-formula">
what is the awnser​
Mathematics
2 answers:
Fittoniya [83]3 years ago
5 0

Answer:

t ≤-3

Step-by-step explanation:

5t≤-15

Divide by 5

5t/5 ≤-15/5

t ≤-3

Rus_ich [418]3 years ago
5 0
T < -3 (put a line underneath the <)
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Answer:

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AB = 20 metres

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Step-by-step explanation:

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A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 0.1
Ghella [55]

Answer:

z=-0.674

And if we solve for \mu we got

\mu=12 +0.674*0.4=12.270

So then the mean is 'mu = 12.270 for this case.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sqrt{0.16}= 0.4)  

For this part we want to find a value a, such that we satisfy this condition:

P(X>12)=0.85   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.85 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.85

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for \mu we got

\mu=12 +0.674*0.4=12.270

So then the mean is 'mu = 12.270 for this case.

8 0
3 years ago
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gtnhenbr [62]

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Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

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You will note that this approach works well enough for ...

  a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

  (x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

  x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0
3 years ago
Please answer 1-5 i will give brainiest and 40 points
sergij07 [2.7K]

Answer:Okay I'm really confused

Step-by-step explanation:

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3 years ago
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<h3>ᎪꪀsωꫀᏒ</h3>

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