PLEASE HELP WILL MARK BRAINEST

Elizabeth is going to invest $480 and leave it in an account for 8 years. Assuming the

son4ous [18]

Answer:4.2%

Step-by-step explanation: it’s what delta math told me the answer was

7 0

3 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

Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is $0.95. The number

forsale [732]

I’m guessing that this is for systems of equations?
n - nickels
d - dimes

These are the equations you start off with.
n = 5d-2
0.95 = 0.10d+0.05n

Substitute the top equation for n into the variable n in the bottom equation.
0.95 = 0.10d+0.05(5d-2)

Solve for d.
0.95 = 0.10d+0.25d-0.10
1.05 = 0.35d
3 = d

Substitute d into the top equation and solve for n.
n = 5(3)-2
n = 15-3
n = 12

There are 3 dimes and 12 nickels in the coin purse! Hope this helped <3

7 0

3 years ago

Read 2 more answers

Please help! 20 points

Bas_tet [7]

Substitute each value in:

x < 3

y > 1

5 < 3

2 > 1

0 < 3

0 > 1