Which expression is equivalent to (4−3)−6? 418 43 4−18 4−9

Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a rando

DENIUS [597]

Answer:

$n=13.167^2 =173.369$ and if we round up to the nearest integer we got n =174

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =73,\sigma =10.5)$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Solution to the problem

We want to find the value of n that satisfy this condition:

$P(71.5 < \bar X$

And we can use the z score formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we have this:

$P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z$

And we can express this like this:

$P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94$

And by properties of the normal distribution we can express this like this:

$P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z$

If we solve for $P(Z$ we got:

$P(Z$

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: $z=-1.881$

And using this we have this equality:

$-1.881 = -0.14286 \sqrt{n}$

If we solve for $\sqrt{n}$ we got:

$\sqrt{n} = \frac{-1.881}{-0.14286}=13.167$

And then $n=13.167^2 =173.369$ and if we round up to the nearest integer we got n =174

6 0

3 years ago

I need help with this

defon

i guess just times them all then divide up by how many numbers you have

3 0

3 years ago

20 PTSS !!! PLEASE HELPP !!

butalik [34]

Because it is extremely hard to find the area of this figure all together, it would be in our best interest to split this figure up into three different pieces: the two horizontal rectangles, and the verticle rectangle. We can find the area of all three and add them up. Be aware that there are two different ways that you can break this figure up, As shown in the attachments. I will be using the first image (the one with the tall horizontal rectangles, NOT the almost-squares).

So, we see that we have enough information to solve for the area of the left-most rectangle. Area = lw. 10 x 4 = 40, so the area is 40. Next, we have to notice, that the horizontal rectangles are also the same, so both of the areas of the two horizontal rectangles are 40.

Now, we can find the middle rectangle. We know that the length of the entire thing is 18, but it is taken up by 8 (4+4) of the horizontal triangles, so 18-8=10, so the length Is 10. We also know that the height of the horizontal rectangles is 10, so 10-3=7. Our dimensions for the rectangle are 10x7 or 70 square units. If we add them all together, 40+40+70=150.

The area is 150 square units

8 0

3 years ago

Multiply this expression :: -2/7 • (-3 5/8)

Helen [10]

Answer: 29/28

Step-by-step explanation:

-2/7 x (-3 5/8)

= -2/7 x -29/8

= 2/7 x 29/8 = 1/7 x 29/4

= 29/28

7 0

3 years ago

Here is a growing pattern of squares:

Flura [38]

Answer:

squares in Step n. f (n) = 8 + 3(n – 1} for n > 1 /(1) = 8, /{n2) = 3+f (n – 1) for n > 2 01)= 8, 7 (n) = 8= ƒ(n=1) forn> 2 Df1)= 3 -8 (n- 1) forn > 1 Of (n) - 37 + 5 for n > 1 32+5 for n>1 CS (n) 3+ an forn 1

Step-by-step explanation:

3 0

2 years ago