One third plus x equals 9

The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location. Find an equiva

kolezko [41]

Answer:

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

Step-by-step explanation:

The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location.

The inequality expression is given as:

$|T-43|\leq 22$

now this expression could also be expressed as:

$-22\leq T-43\leq22\\\\-22+43 \leq T \leq 22+43\\\\21\leq T\leq 65$

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

( As the difference is 22 degrees to the left and right)

5 0

3 years ago

Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose

spayn [35]

Answer:

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

The home team therefore wins 50% of its games

This means that $p = 0.5$

Determine the probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Sample of 80 means that $n = 80$ and, by the Central Limit Theorem:

$\mu = p = 0.65$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{80}} = 0.0559$

This probability is 1 subtracted by the pvalue of Z when X = 0.65. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.65 - 0.5}{0.0559}$

$Z = 2.68$

$Z = 2.68$ has a pvalue of 0.9963

1 - 0.9963 = 0.0037

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

7 0

2 years ago

Please help! 25points! pleaseeee helppppp

BaLLatris [955]

Answer: this is difficult

Step-by-step explanation:

6 0

2 years ago

At the beginning of an environmental study, a forest covered an area of 1500 km2 . Since then, this area has decreased by 9.8% e

fredd [130]

Answer:

$A(t)=(0.902)^t \cdot 1500$ $[km^2]$

Step-by-step explanation:

In this problem, the initial area of the forest at time t = 0 is

$A_0 = 1500 km^2$

After every year, the area of the forest decreases by 9.8%: this means that the area of the forest every year is (100%-9.8%=90.2%) of the area of the previous year.

So for instance, after 1 year, the area is

$A_1 = A_0 \cdot \frac{90.2}{100}=0.902 A_0$

After 2 years,

$A_2=0.902 A_1 = 0.902(0.902A_0)=(0.902)^2 A_0$

And so on. So, after t years, the area of the forest will be

$A(t)=(0.902)^t A_0$

And by substituting the value of A0, we can find an explicit expression:

$A(t)=(0.902)^t \cdot 1500$ $[km^2]$

8 0

3 years ago

Which equation represents a parabola opening to the right with a vertex at the origin and focus at (9,0) A. X = - 1/36y^2 B. X =

Virty [35]

Answer:

Option C: x = (1/36)*y^2

Step-by-step explanation:

A horizontal parabola is written as

x = f(y) = a*y^2 + b*y + c

If a is positive, the parabola opens to the right.

if a is negative, the parabola opens to the left.

And the vertex of the parabola has the y-value:

y = -b/2a

and the x value

x = f(-b/2a)

For our parabola, we know that:

Opens to the right:

Then the only options left are:

C) x = (1/36)*y^2

D) x = (1/6)*y^2

Because in both cases b = 0, both of the equations have the vertex in the point (0, 0).

Now let's see wich one has a focus at (9, 0)

If the vertex of our equation is:

(h, k)

Then the focus will be:

(h + a, k)

Where a is the directrix of the equation.

Here we know that the vertex is (0, 0) and the focus is (9, 0)

Then:

(0 + a, 0) = (9, 0)

The directrix is 9.

And the directrix is such that:

(x - h)^2 = 4*a*(y - k)

Replacing the values of h and k (both are zero) we get:

x^2 = 4*a*y

And we know that: a = 9

x^2 = 4*9*y

x^2 = 36*y

if we isolate y, we get:

(1/36)*x^2 = y

This is option C.

Then the correct option is C.

8 0

2 years ago