Notebooks are originally on sale for $2.00. This school year they were marked up by 15%. What's

24 students of 30 is what percent

____ [38]

The answer is 80 percent

5 0

2 years ago

Probability can be calculated using the formula

AveGali [126]

It really depends on the situation as probability depends on logic more than simple mathematical equations. However you must keep in mind a few principles:
1- P = number of possible outcomes of the event / total outcomes
2- P always less than or equal to 1.
3- For independent events A and B to occur simultaneously, P= P(A) x P(B)

You can send me the question you’re stuck at for further help.

6 0

3 years ago

What is the equation of a sine function with an amplitude of 2 and a period of 4pi.

Setler [38]

Answer:

um

Step-by-step explanation:

7 0

2 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

Life rating in Greece. Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sa

elena-s [515]

Answer:

a

The population parameter of interest is the true proportion of Greek who are suffering

While the point estimate of this parameter is proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%

b

The condition is met

c

The 95% confidence interval is $0.223 < p < 0.277$

d

If the confidence level is increased which will in turn reduce the level of significance but increase the critical value( $Z_{\frac{\alpha }{2} }$ ) and this will increase the margin of error( deduced from the formula for margin of error i.e $E \ \alpha \ Z_{\frac{\alpha }{2} }$ ) which will make the confidence interval wider

e

Looking at the formula for margin of error if the we see that if the sample size is increased the margin of error will reduce making the confidence level narrower

Step-by-step explanation:

From the question we are told that

The sample size is n = 1000

The population proportion is $\r p = 0.25$

Considering question a

The population parameter of interest is the true proportion of Greek who are suffering

While the point estimate of this parameter is proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%

Considering question b

The condition for constructing a confidence interval is

$n * \r p > 5\ and \ n(1 - \r p ) >5$

So

$1000 * 0.25 > 5 \ and \ 1000 * (1-0.25 ) > 5$

$250 > 5 \ and \ 750> 5$

Hence the condition is met

Considering question c

Given that the confidence level is 95% then the level of significance is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_\frac{ \alpha }{2} * \sqrt{ \frac{\r p (1 - \r p ) }{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{ 0.25 (1 - 0.25 ) }{ 1000} }$

$E = 0.027$

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.25 - 0.027 < p < 0.25 + 0.027$

substituting values

$0.223 < p < 0.277$

considering d

If the confidence level is increased which will in turn reduce the level of significance but increase the critical value( $Z_{\frac{\alpha }{2} }$ ) and this will increase the margin of error( deduced from the formula for margin of error i.e $E \ \alpha \ Z_{\frac{\alpha }{2} }$ ) which will make the confidence interval wider

considering e

Looking at the formula for margin of error if the we see that if the sample size is increased the margin of error will reduce making the confidence level narrower

5 0

3 years ago