The difference of c and 13 is less than -19.

Rewrite y =<img src="https://tex.z-dn.net/?f=%5Csqrt9%7Bx%7D%20%2B45" id="TexFormula1" title="\sqrt9{x} +45" alt="\sqrt9{x} +45"

tigry1 [53]

Answer:

(1) We have been given a function: $y=\sqrt{9x}+45-2$

We can rewrite it using translation is:

It is the graph of $y=\sqrt{9x}$ translated 45 units left and 2 units down.

(2) Now, $y=\sqrt{x}-5-2$

So this is the graph of $y=\sqrt{x}$ translated 5 units right and 2 units down.

(3) $y=\sqrt{x}+5-2$

So, this is the graph of $y=\sqrt{x}$ translated 5 units left and 2 units down.

(4) $y=\sqrt[3]{x}+5-2$

It is the graph of $y=\sqrt[3]{x}$

translated 5 units left and 2 units down.

(5) $y=\sqrt[3]{x}+5-2$

It is the graph of $y=\sqrt[3]{x}$

translated 5 units left and 2 units down.

8 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

What is the area of a rectangle with vertices at (−3, −1) , (1, 3) , (3, 1) , and (−1, −3) ?

77julia77 [94]

The coordinates of the rectangle are (-3,-1), (1,3), (3,1) and (-1,-3).

Draw the given rectangle as shown in the figure below.

Calculate the length of the rectangle as
a = √(4² + 4²) = 4√2
Calculate the width of the rectangle as
b = √(2² + (-2)²) = 2√2
Calculate the area of the rectangle as
A = a*b = (4√2) * (2√2) = 16

Answer: 16

4 0

3 years ago

The slope of the normal curve y=3x^2-5x is at the point (0,0)

Rufina [12.5K]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Consider the curve $y=3x^2-5x$

Differentiate with respect to $x$

$\frac{dy}{dx} =3(2x)-5=6x-5$

( Use $\frac{d}{dx}(x^n)=nx^{n-1}$ )

Slope of normal curve is given by $\frac{-1}{\frac{dy}{dx} }$

Slope of normal curve = $\frac{-1}{6x-5}$

To find slope of normal curve at point $(0,0)$ ,put $x=0$ in $\frac{-1}{6x-5}$

Slope of normal curve at point $(0,0)$ = $\frac{-1}{6(0)-5} =\frac{-1}{-5}=\frac{1}{5}$

8 0

3 years ago

Does anymore know? Thennddndn

OverLord2011 [107]

Answer:

C

Step-by-step explanation:

not enough information to prove they are congruent. the postulates need 3 things that are congruent, unless it's a right triangle, you only need two. However, in this image, they are not shown to be right triangles.

8 0

3 years ago