What does it mean when a function is one-to-one?

1 answer:

3 0

Let a function be defined from A to B then f is said to be one to one function if every element of domain has its unique image in co- domain . In other words, if each separate elements of set A are mapped with separate of set B then such function is known as one to one function.

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hammer [34]

Answer:

The test statistic is z=2.5.

The null hypothesis is rejected.

There is enough evidence to support the claim that more than 10% of the 16 ounce cups are underfilled.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that more than 10% of the 16 ounce cups are underfilled.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.1\\\\H_a:\pi>0.1$

The significance level is assumed to be 0.05.

The sample has a size n=100.

The sample proportion is p=0.18.

$p=X/n=18/100=0.18$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.1*0.9}{100}}\\\\\\ \sigma_p=\sqrt{0.001}=0.03$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.18-0.1-0.5/100}{0.03}=\dfrac{0.075}{0.03}=2.5$

This test is a right-tailed test, so the P-value for this test is calculated as:

$P-value=P(z>2.5)=0.006$

As the P-value (0.006) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that more than 10% of the 16 ounce cups are underfilled.

7 0

3 years ago

The pressure and volume of an expanding gas are related by the formula PV^b=C, where b and C are constants (this holds in adiaba

bazaltina [42]

The dP/dt of the adiabatic expansion is -42/11 kPa/min

<h3>How to calculate dP/dt in an adiabatic expansion?</h3>

An adiabatic process is a process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression

Given b=1.5, P=7 kPa, V=110 cm³, and dV/dt=40 cm³/min

PVᵇ = C

Taking logs of both sides gives:

ln P + b ln V = ln C

Taking partial derivatives gives:

$\frac{1}{P}\frac{∂P}{∂t} + \frac{b}{V}\frac{∂V}{∂t} = 0$