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Ronch [10]
3 years ago
5

If $9x^2 - 16x + k$ is a perfect square trinomial, find $k$.

Mathematics
1 answer:
Lady_Fox [76]3 years ago
7 0

Answer:is this a real quistion

Step-by-step explanation:?

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If Upper X overbar equals 62​, Upper S equals 8​, and n equals 36​, and assuming that the population is normally​ distributed, c
marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)    

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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=62 represent the sample mean

\mu population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size  

Part a: 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=36-1=35

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that t_{\alpha/2}=2.72

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

62-2.72\frac{8}{\sqrt{36}}=58.373    

62+2.72\frac{8}{\sqrt{36}}=65.627    

So on this case the 99% confidence interval would be given by (58.373;65.627)    

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

3 0
3 years ago
Not tryna be weird or nun but if anyone would like to join my g0ogle classroom for help with advice for teens from teens AGES 13
boyakko [2]

Answer:

ight

Step-by-step explanation:

I'll think bout it.

I may join, but I won't be on that much.

6 0
3 years ago
What is the slope of a horizontal line
Step2247 [10]

Answer:

0

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4
arsen [322]

we are given

R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k

now, we can find x , y and z components

x=cos(8t),y=sin(8t),z=8ln(cos(t))

Arc length calculation:

we can use formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

x'=-8sin(8t),y=8cos(8t),z=-8tan(t)

now, we can plug these values

L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt

now, we can simplify it

L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8sec(t) dt

now, we can solve integral

\int \:8\sec \left(t\right)dt

=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|

now, we can plug bounds

and we get

=8\ln \left(\sqrt{2}+1\right)-0

so,

L=8\ln \left(1+\sqrt{2}\right)..............Answer

5 0
2 years ago
Up arrow, right arrow, down arrow, left arrow, up arrow, right arrow, down arrow, left arrow, what’s the conjunction
AlladinOne [14]

Step-by-step explanation:

↑

→

↓

←

↑

→

↓

←

conjunction are the words that join 2 sentences

8 0
2 years ago
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