See the picture attached to better understand the problem
we know that
If two secant segments are drawn to a <span>circle </span><span>from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
</span>so
jl*jk=jn*jm------> jn=jl*jk/jm
we have
<span>jk=8,lk=4 and jm=6
</span>jl=8+4----> 12
jn=jl*jk/jm-----> jn=12*8/6----> jn=16
the answer isjn=16
The solution to the problem is as follows:
<span>Area=h×w
</span><span>1(1/2) = h × 2 (<span>1/4)
</span></span><span>⟹3/2 = h × <span>9/4
</span></span><span>h = 3/2 ÷ <span>9/4
</span></span><span>Therefore, the height will be 2/3 yards if we use all the blue paint.
</span>
I hope my answer has come to your help. Have a nice day ahead and may God bless you always!
Answer:
x^2 + 1.1x + 0.28
Step-by-step explanation:
<span>3x = y + 7
</span><span> y =3x-7
</span>
<span>8x = 2y + 5
</span><span>8x = 2(3x-7) + 5
8x=6x-14+5
8x-6x=-9
2x=-9
x=-4.5
</span><span>8x = 2y + 5
</span>8(4.5)=2y+5
36=2y+5
2y=31
y=15.5
We assume the lunch prices we observe are drawn from a normal distribution with true mean 
and standard deviation 0.68 in dollars.
We average 
samples to get 
.
The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write
Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains .
Our interval takes the form of 
as is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".
Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.
Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.
With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.
We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is
in other words a margin of error of
dollars
That's around plus or minus 17 cents.
