What is the final answer

How many times can 1000000000000 go into 10

makkiz [27]

Answer:

10 million (10,000,000) tens can go into 100,000,000

Step-by-step explanation:

4 0

3 years ago

ShoeShop is having a sale on flip-flops. The first 2 pairs cost $8.50 each plus each additional pair costs $3.75. Mrs. Tyler can

olga55 [171]

No matter what, if you get at least 2 pair of flip flops you will be spending $8.50. So that's the amount you're going to spend plus 3.75 per pair after that. If she has $34.75 to spend and cannot go over, the inequality sign will be less than or equal to. It's ok if she spends less than that and it's ok if she spends exactly that, but she cannot spend more than she has. The inequality then would be this: $3.75p+8.50 \leq 34.75$

7 0

3 years ago

Read 2 more answers

Make w the subject of the formula s = w/a

enot [183]

Answer:

w = sa

Step-by-step explanation:

Given

s = $\frac{w}{a}$ ( multiply both sides by a to clear the fraction )

sa = w

8 0

3 years ago

8.25 halfway to the nearest tenth

Leya [2.2K]

Answer:

10

Step-by-step explanation:

4 0

4 years ago

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com

Ymorist [56]

Answer:

a) $183-1.984\frac{20}{\sqrt{100}}=179.032$

$183+1.984\frac{20}{\sqrt{100}}=186.968$

So on this case the 95% confidence interval would be given by (179.032;186.968)

b) $190-1.984\frac{23}{\sqrt{100}}=185.437$

$190+1.984\frac{23}{\sqrt{100}}=194.563$

So on this case the 95% confidence interval would be given by (185.437;194.563)

c) For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

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".

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

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Part a : Summer

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:

$df=n-1=100-1=99$

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(0.025,99)".And we see that $t_{\alpha/2}=1.984$

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

$183-1.984\frac{20}{\sqrt{100}}=179.032$

$183+1.984\frac{20}{\sqrt{100}}=186.968$

So on this case the 95% confidence interval would be given by (179.032;186.968)

Part b: Winter

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:

$df=n-1=100-1=99$

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(0.025,99)".And we see that $t_{\alpha/2}=1.984$

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

$190-1.984\frac{23}{\sqrt{100}}=185.437$

$190+1.984\frac{23}{\sqrt{100}}=194.563$

So on this case the 95% confidence interval would be given by (185.437;194.563)

Part c

For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

5 0

3 years ago