Need help to solve question

Here is the histogram of a data distribution.

Doss [256]

Answer:

Uniform

Step-by-step explanation:

We don't have any peaks in the histogram, so its neither unimodal or bimodal leaving only uniform as the answer.

3 0

3 years ago

Find the missing quotient ? ____ 30( 240

Arada [10]

Answer:

8

Step-by-step explanation:

30( 240=8

30x9=240

Have amazing day!

6 0

3 years ago

So im really confused by this and could use some help.

Otrada [13]

Answer:

Nancy fits the equation.

Step-by-step explanation:

I think they're asking which gorilla fits the equation.

2b - 15 = 605

Add 15 to both sides

2b=620

Divide both sides by 2

b=310

The only gorilla who weighs 310 pounds is Nancy.

5 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

(Please help)Joe has a savings account and a credit card. He has $422 saved in his savings account and has purchased $524 worth

ruslelena [56]

He now has -102 dollars

422-524= -102

6 0

3 years ago