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3 years ago

Choose the equation that cannot be used to find the

Mathematics

1 answer:

vekshin13 years ago

6 0

I believe it’s the 3rd option

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Uscles Gym has a membership fee of $40 every 3 months. Energy Gym has a membership fee of $93 every 6 months.

snow_lady [41]

Answer:

Muscles Gym: 13.33

Energy Gym: 15.00

Step-by-step explanation:

Muscles Gym: You divide "40.00" by "3", which will then give you the answer of "13.33".

Energy Gym: You divide "93.00" by "6", which will then give you the answer of "15.50".

You then add the zero to show the proper number of cents after the decimal point.

I hope this helps!

6 0

3 years ago

4. According to statistics reported on IN-CORP a surprising number of motor vehicles are not covered by insurance. Sample result

son4ous [18]

Answer:

a) 0.23

b) The 95% confidence interval for the population proportion is (0.1717, 0.2883).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Point estimate

The point estimate is:

$\pi = \frac{46}{200} = 0.23$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.23 - 1.96\sqrt{\frac{0.23*0.77}{200}} = 0.1717$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.23 + 1.96\sqrt{\frac{0.23*0.77}{200}} = 0.2883$

The 95% confidence interval for the population proportion is (0.1717, 0.2883).

5 0

3 years ago

Solve the equation. 3|3k| = 27

Ratling [72]

The answer is B. -3, 3
This is why:
3|3*3| = 27
3|9| = 27
3*9 = 27
The same will happen with -3 since this | means the distance from the number to zero and in this case it is 3 units til you reach zero. So the answer is B. 3, -3

3 0

3 years ago

A dartboard has a diameter of 30 inches. What are its radius and circumference?

inessss [21]

Answer:

radius = 15 in.

circumference = 94.2 in.

Step-by-step explanation:

radius = diameter/2 = 30 in./2 = 15 in.

circumference = (pi)d = 3.14 * 30 in. = 94.2 in.

4 0

2 years ago

F(x) = (128/127)(1/2)x, x = 1,2,3,...7. determine the requested values: round your answers to three decimal places (e.g. 98.765)

Marrrta [24]

A.
$\mathbb P(X\le 1)=\mathbb P(X=1)=\dfrac{128}{127}\left(\dfrac12\right)^1=\dfrac{64}{127}$

b.
$\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\dfrac{64}{127}=\dfrac{63}{127}$

c.
$\mathbb E(X)=\displaystyle\sum_{x=1}^7 x\,f_X(x)=\frac{64}{127}\sum_{x=1}^7 x\left(\frac12\right)^{x-1}$

Suppose $f(y)=\displaystyle\sum_{x=0}^7 y^x$ . Then $f'(y)=\displaystyle\sum_{x=1}^7 xy^{x-1}$ . So if we can find a closed form for $f(y)$ , in terms of $y$ , we can find $\mathbb E(X)$ by evaluating the derivative of $f(y)$ at $y=\dfrac12$ .

$f(y)=\displaystyle\sum_{x=0}^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7$
$y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8$
$f(y)-y\,f(y)=y^0-y^8$
$(1-y)f(y)=1-y^8$
$f(y)=\dfrac{1-y^8}{1-y}$
$\implies f'(y)=\dfrac{7y^8-8y^7+1}{(1-y)^2}$
$\implies\mathbb E(X)=\dfrac{64}{127}f'\left(\dfrac12\right)=\dfrac{64}{127}\times\dfrac{247}{64}=\dfrac{247}{127}$

d.
$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

We find $\mathbb E(X^2)$ in a similar manner as in (c).

$\mathbb E(X^2)=\displaystyle\sum_{x=1}^7 x^2\,f_X(x)=\frac{32}{127}\sum_{x=1}^7x^2\left(\frac12\right)^{x-2}$

Now,

$f(y)=\displaystyle\sum_{x=0}^7y^x$
$\implies f'(y)=\displaystyle\sum_{x=1}^7xy^{x-1}$
$\implies f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$

We know that

$f''(y)=-\dfrac{42y^8-96y^7+56y^6-2}{(1-y)^3}$
$\implies f''\left(\dfrac12\right)=\dfrac{219}{16}$

We also have

$f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$
$f''(y)=\displaystyle\sum_{x=2}^7x^2y^{x-2}-\sum_{x=2}^7xy^{x-2}$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=2}^7x^2y^x-\sum_{x=2}^7xy^x\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\bigg(\sum_{x=1}^7x^2y^x-y\bigg)-\bigg(\sum_{x=1}^7xy^x-y\bigg)\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=1}^7x^2y^x-\sum_{x=1}^7xy^x\right)$

so that when $y=\dfrac12$ , we get

$\dfrac{219}{16}=4\left(\dfrac{127}{128}\mathbb E(X^2)-\dfrac{127}{128}\mathbb E(X)\right)\implies\mathbb E(X^2)=\dfrac{685}{127}$

Then

$\mathbb V(X)=\dfrac{685}{127}-\left(\dfrac{247}{127}\right)^2=\dfrac{25,986}{16,129}$

6 0

3 years ago