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artcher [175]
3 years ago
10

Can you Solve. 3x^2+10x+3

Mathematics
2 answers:
Alika [10]3 years ago
6 0
You cannot solve for x but you can factor

to solve in ax^2+bx+c form you must find
b=x+y
a times c=x times y

so solve

3 times 3=9

what 2 number multily to get 9 and add to get 10
9=x times y
the numbers are 9 and 1 so

3x^2+1x+9x+3
(3x^2+1x)+(9x+3)
factor
(x)(3x+1)+(3)(3x+1)
reverse distribute
ab+ac=a(b+c)

(x)(3x+1)+(3)(3x+1)=(x+3)(3x+1)
factored out form is (x+3)(3x+1)

sp2606 [1]3 years ago
4 0
<span>Simplifying
3x^2 + 10x + 3
 Reorder the terms
3 + 10x + 3x^2
Factor a trinomial
(3 + x)(1 + 3x)
Final result
(3 + x)(1 + 3x)</span>
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The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m
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Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

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Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

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And we have;

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Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of \bar X Round your answers to two decimal places.

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 delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

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