Find the sum of the polynomials below.

A child lies on the ground and looks up at the top of a 9-ft tree nearby. The child is 13 ft away from the tree. What is the ang

choli [55]

Draw a right triangle to represent the problem.
The vertical height of the triangle is 9 ft, and it represents the tree.
The horizontal length, at the bottom of the tree is ground level and has a length of 13 ft.

Let x = angle of elevation.

By definition,
tan x = 9/13 = 0.6923
x = arctan(0.6923) = 34.7 deg. = 35 deg (approx)

Answer: 35°

8 0

2 years ago

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

Suppose you pick two cards from a deck of 52 playing cards. What is the probability that they are both queens?

photoshop1234 [79]

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

$D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6$