56*10<br><br> 3682*356<br><br> 409*53793

-5.73q = 97.41 q= please help me???

jek_recluse [69]

First Divide both sides by -5.73

-5.73q/-5.73=97.41/-5.73

Answer: q=-17

7 0

3 years ago

A time series that shows a recurring pattern over one year or less is said to follow a Select one:______

zaharov [31]

The answer is C. Seasonal pattern

3 0

3 years ago

Which expression is equivalent to - (4-3)

LenaWriter [7]

This is equivalent to -1 ( 4 - 3 )
which simplified is -1
because 4-3 is 1 and 1 times -1 is -1

Hope this helps ;)

5 0

3 years ago

Read 2 more answers

Required information An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiv

Natasha2012 [34]

Answer:

$652.6-2.01\frac{311.7}{\sqrt{50}}=563.997$

$652.6+2.01\frac{311.7}{\sqrt{50}}=741.203$

So on this case the 95% confidence interval would be given by (563.997;741.203)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=652.6$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=311.7 represent the sample standard deviation

n=50 represent the sample size

Soltuion to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=50-1=49$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$