What is 25% of 15 Please help thank you!

A value meal package at Ron's Subs consists of a drink, a sandwich, and a bag of chips. There are 4 types of drinks to choose fr

Temka [501]

To answer this question, multiply all given numbers together.

4*3*3

12*3

36

36 different value meal packages are possible.

Hope this helps!

8 0

3 years ago

Read 2 more answers

An armount of nitrogen undergoes a physical change and increases its volurme, What can be knoW) about this

Allushta [10]

Answer:

d

Step-by-step explanation:

8 0

3 years ago

Carol is cross country skiing the table shows the distance traveled after various numbers of minutes what is the rate of change

MatroZZZ [7]

Answer:

Option D 3/16

Step-by-step explanation:

Carol is cross-country skiing.

With the help of the given table we have to calculate the rate of change.

If we draw a graph for distance traveled on y-axis and tins at x-axis we find two points, ( 2, 1/6) and (3, 17/48).

Then slope of the line connecting these points will be the rate of change.

8 0

3 years ago

Lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. a bank conducts inter

Otrada [13]

Part A:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85

Thus, the probability that the lie detector will conclude that all 15 are telling the truth if all 15 applicants tell the truth is given by:

 $P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\ \\ =1\times0.0874\times1=0.0874$


Part B:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15

Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:

$P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\ \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\ \\ =0.9126$

Part C:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The mean is given by:

$\mu=npq \\ \\ =15\times0.15\times0.85 \\ \\ =1.9125$

Part D:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The probability that the number of truthful applicants classified as liars is greater than the mean is given by:

 $P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\ \\ 1-[P(0)+P(1)]$

 $P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\ \\ =15\times0.15\times0.1028=0.2312$

8 0

3 years ago

A researcher claims that the average annual salary of part-time community college instructors is at least $45,000. To test this

Doss [256]

Answer:

See explanation

Step-by-step explanation:

Solution:-

- The researcher claims :

" the average annual salary of part-time community college instructors is at least $45,000 "

- We will extract the claim made by the researcher to be the Null hypothesis:

Null Hypothesis : μ ≥ $45,000

- Any statement otherwise, or point of rejection criteria will denote the Alternate hypothesis that disbands the claim that:

" the average annual salary of part-time community college instructors is less than $45,000 " :

Alternate Hypothesis : μ < $45,000

- This type of test with a sample size of n = 25 < 30 and population standard deviation is also not given hints the use t-test statistics and t-critical value reject the claim.

Lower tail - one sample - T-test.

6 0

3 years ago