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2 years ago

How do you write 2.59 x 10-3 in standard form?

Mathematics

1 answer:

Allisa [31]2 years ago

7 0

Answer:

2.59 x 10-3 = 22.9

and 22.9 in standard form is Your number written in expanded form is:

20 + 2 + 0.9

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Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

laila [671]

Answer:

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

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".

$\bar X=17$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

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:

$df=n-1=10-1=9$

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(0.025,9)".And we see that $t_{\alpha/2}=2.262$

Now we have everything in order to replace into formula (1):

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

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3 years ago