All categories

3 years ago

Can someone help me please

Mathematics

2 answers:

Alex17521 [72]3 years ago

8 0

Answer:

D. The mode for sunday is 15

Step-by-step explanation:

Gnom [1K]3 years ago

3 0

Answer:

The mode of sunday is 15 is correct

You might be interested in

Pls help. Cindi wants to cover the top and sides of this box with glass tiles that

Marat540 [252]

Answer:

300 square centimeters

4 0

3 years ago

Read 2 more answers

jackson is going to the movies. he has to pay $7 for his ticket and $2 for each box of candy he buys. write an expression that r

ruslelena [56]

If Jackson went to the movies and payed $7 initially for his ticket and then payed $2 for each box of candy that he buys, then the equation could look like this: y = 2x + 7, where y represents the total cost and x represents the number of boxes of candy Jackson buys.

6 0

4 years ago

Read 2 more answers

Can a -150/10 be classified as both a rational number and a integer

irakobra [83]

Yes! This is because -150/10 can be simplified to be -15, which is a rational number.

The word “rational” sounds like another math word you’ve heard of before. Do you know what it is?

Well, it’s “ratio”!! Ratios can be seen in the forms x:y and x/y.

ANY RATIONAL NUMBER HAS THE ABILITY TO BE WRITTEN AS A RATIO!! This will completely exclude numbers with super long decimal points (ex: 1.2345678809928374737272828...)

This number also meets the requirements of being an integer. An integer is any whole number (this excludes decimals and fractions)

I know it’s written as a fraction. However, the fraction could be simplified, making it -15, which means this is both a rational number and an integer!!

6 0

3 years ago

Help with both questions offering 10 points

Alexandra [31]

Answer:

$\:\mathrm{First\: problem:\:}-2,\\\:\mathrm{Second\: problem:\: }-3$

Step-by-step explanation:

The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$

Plugging in any two points in the table we have:

First problem:

$m=\frac{1-5}{2-0}=\frac{-4}{2}=\fbox{$-2$}$

Second problem:

$m=\frac{-1-5}{2-0}=\frac{-6}{2}=\fbox{$-3$}$

8 0

3 years ago

Read 2 more answers

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

4 years ago