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3 years ago

Can someone explain how to do this? thanksss <33

Mathematics

1 answer:

Ymorist [56]3 years ago

4 0

Answer:

True True False True

Step-by-step explanation:

You would substitute the value for x and the value for y into both equations and see if the sign holds true.

So for number 1:

x=3, y=-2;

y ≤ x -3

-2 ≤ x - 3

-2 ≤ -1

y ≥ -x - 2

-2 ≥ -3 - 2

-2 ≥ -5

Number 1 is true because the equations are true ^^

Do this for every problem so it would be:

1. True

2. True

3. False

4. True

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The second statement is the what of the first?

Tamiku [17]

Answer:

contrapositive

Step-by-step explanation:

The converse statement would be ...

b ⇒ a

The contrapositive swaps the parts and negates them both:

¬b ⇒ ¬a . . . . . matches your given statement

The inverse negates both parts:

¬a ⇒ ¬b

A contradiction arises when the implication is false: a is true and b is false.

6 0

4 years ago

Assume that the last two digits on a car number plate are equally likely to be any of the one hundred outcomes {00, 01, 02, ....

gizmo_the_mogwai [7]

Answer:

1) Probability that Peter wins = PeW = 48.53%, while the probability that Paul wins = PaW= 1-PeW = 51.47% (Peter si favoured)

2) The bet will be pretty fair at n=16 or 17

Step-by-step explanation:

Since each car number plate is independent of the others, then the random variable X=number of cars that have the same last two digits follows a binomial distribution. Thus the probability distribution is

P(X) = n!/((n-x)!*x!)*p^n*(1-p)^(n-x)

where

p= probability that a car has the same last 2 digits = {00, 11, ...,99}/{00, 01, 02, ........ 98, 99} = 10/100 = 1/10

n= number of cars = 16

Then the probability that Peter wins PeW is:

PeW=P(X≥2) = 1- P(X<2) = 1- F(X=2) = 1 - 0.5147 = 0.4853 = 48.53%

therefore

Probability that Peter wins = PeW = 48.53%

Probability that Paul wins = PaW= 1-PeW = 51.47% (Peter si favoured)

for n=17 , PeW = 0.518 , PaW= 0.482

for n=15 , PeW = 0.450 , PaW= 0.548

then the bet will be pretty fair at n=16 or 17

6 0

3 years ago

The diameters of ball bearings are distributed normally. The mean diameter is 106 millimeters and the standard deviation is 4 mi

Assoli18 [71]

Answer:

Probability that the diameter of a selected bearing is greater than 111 millimeters is 0.1056.

Step-by-step explanation:

We are given that the diameters of ball bearings are distributed normally. The mean diameter is 106 millimeters and the standard deviation is 4 millimeters.

Firstly, Let X = diameters of ball bearings

The z score probability distribution for is given by;

Z = $\frac{ X - \mu}{\sigma}$ ~ N(0,1)