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3 years ago

PLEASE HELP!! Determine whether f is an exponential function of x. If so, find the constant ratio.

Mathematics

1 answer:

yaroslaw [1]3 years ago

8 0

Answer: Yes f is a function of x

Step-by-step explanation: f(x) means fuctoin of x thats what it stands for.

and it has a constant ratio of 1/2. Hope this helps!!

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PLS HELP ASAPPPPP WITH PART B, C, & D PLS PLS

Westkost [7]

Answer:

Part b would be 4 and c would be 5 and d would be 19

Step-by-step explanation:

So yea do those

3 0

3 years ago

A bowling leagues mean score is 197 with a standard deviation of 12. The scores are normally distributed. What is the probabilit

Bond [772]

Answer:

0.281 = 28.1% probability a given player averaged less than 190.

Step-by-step explanation:

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

A bowling leagues mean score is 197 with a standard deviation of 12.

This means that $\mu = 197, \sigma = 12$

What is the probability a given player averaged less than 190?

This is the p-value of Z when X = 190.

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

$Z = \frac{190 - 197}{12}$

$Z = -0.58$

$Z = -0.58$ has a p-value of 0.281.

0.281 = 28.1% probability a given player averaged less than 190.

8 0

3 years ago

A spinner has 8 sections. 2 are grey 6 are blue. The spinner is spun twice what is the probability that the first lands on grey

sweet [91]

Answe

Step-by-step explanation:

5 0

3 years ago

Look at the formula for the area of a circle below. . . A = 3.14(pi)r2 . . Which of the following equations can be used to solve

zaharov [31]

<span>You did not include the equations that you want to assess whether they can be used to solve for the radius (r). Likely, the equation of the circumference, C = 2*Pi*r is included, if so => r = C / (2*Pi). If you round Pi to 3.14, the equation may be written r = C / 6.28.</span>

6 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago