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r-ruslan [8.4K]
3 years ago
9

Conditional statements I really need some help

Mathematics
1 answer:
Sloan [31]3 years ago
6 0

p -> q is represented as "If you have a library card, then you can check out books"

~q -> ~p is represented as "If you cannot check out books, then you do not have a library card" This is contrapositive by definition.

q -> p is represented as "If you can check out books, then you have a library card" This is the converse

~p -> ~q is represented as "If you do not have library card, then you cannot check out books". This is the inverse

p <-> q is represented as "You have library card if and only if you can check out books". This is the biconditional.


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(NO FILES OR LINKS OR RANDOM WORDS)
ehidna [41]

Answer:

Randomly selecting a six of diamonds - 1 / 52

Randomly selecting a 7, 8, 9 or 10 - 4 / 13

Step-by-step explanation:

There is only 1 six of diamonds in a standard deck of cards. There are 52 cards in a deck, thus the probability of pulling a six of diamonds is 1 in 52.

There are 4 of each card in a deck. so they are 4 7's, 4 8's. 4 9's and 4 10's. And there are a total of 52 cards in a deck. So the probability of pulling a 7,8,9 or 10 are 4 + 4 + 4 + 4 in 52

4 + 4 + 4 + 4 = 16

16 / 52 simplified is 4 / 13 Therefore the is a 4 in 13 chance of pulling a 7 8 9 or 10

The other ones are correct

3 0
3 years ago
1. Write a quadratic function, in standard form, that fits the set of points. Solve it as a system of three equations
sasho [114]

Answer:

The equation is y=2x^2+4x-7

Step-by-step explanation:

Let the quadratic function be


y=ax^2+bx+c


The point (-4,9) must satisfy this function,

\Rightarrow a(-4)^2+b(-4)+c=9


\Rightarrow 16a-4b+c=9...(1)


The point (0,-7) must also satisfy this function,

\Rightarrow a(0)^2+b(0)+c=-7


\Rightarrow c=-7...(2)



The point (1,-1) must also satisfy this function,

\Rightarrow a(1)^2+b(1)+c=-1


\Rightarrow a+b+c=-1...(3)


We put equation 2 into equation 1 to get;


\Rightarrow 16a-4b-7=9


\Rightarrow 16a-4b=16


\Rightarrow 4a-b=4...(5)


We again put equation 2 into equation 3 to get;


\Rightarrow a+b-7=-1


\Rightarrow a+b=6...(6)


We add equation 5 and 6 to get;

5a=10


\Rightarrow a=2


We put a=2 into equation 6 to get;


2+b=6


\Rightarrow b=6-2


\Rightarrow b=4


The equation is therefore y=2x^2+4x-7

5 0
3 years ago
There is a 70% chance that your car will get stuck in the snow during the next big snow fall. Given that you are already stuck i
nikdorinn [45]

Answer:

63%

Step-by-step explanation:

This is a problem of conditional probability.

The two events that are given are:

  • Car stuck in the snow - Let it be event S. P(S) = 70% = 0.70
  • Require a tow truck - Let it be event T.

We have to find the probability of being stuck in the snow AND requiring a tow truck which can be given as P(S and T)

We are also given the conditional probability, which is P(T | S) = 90% = 0.90

Using the given formula for our case we can modify the formula as:

P(T|S)=\frac{P(S \cap T)}{P(S)}

0.90=\frac{P(S \cap T)}{0.70}\\\\ P(S \cap T)=0.90 \times 0.70\\\\ P(S \cap T)=0.63

Therefore, there is 63% (0.63) chance that you will get stuck in the snow with your car AND require a tow truck to pull you out

4 0
3 years ago
A rectangular portrait is 1 meter wide and 1 meter high. It costs $28.27 per meter to put a gold frame around the portrait. How
Karo-lina-s [1.5K]
The cost will be $113.08.
4 0
3 years ago
Read 2 more answers
Determine the equation of the line that is perpendicular to the lines r​(t) equalsleft angle6​t,1 plus3 ​t,4tright angle and R​(
Lynna [10]

\vec r(t)=\langle6t,1+3t,4t\rangle

\vec R(s)=\langle2+s,-8+3s,-12+4s\rangle

Take the derivatives of each to get the tangent vectors:

\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle6,3,4\rangle

\dfrac{\mathrm d\vec R(s)}{\mathrm ds}=\langle1,3,4\rangle

Take the cross product of the tangent vectors to get a vector that is normal to both lines:

\langle6,3,4\rangle\times\langle1,3,4\rangle=\langle0,-20,15\rangle

The two given lines intersect when \vec r(t)=\vec R(s):

\langle6t,1+3t,4t\rangle=\langle2+s,-8+3s,-12+4s\rangle\implies t=1,s=4

that is, at the point (6, 4, 4).

The line perpendicular to both of the given lines through the origin is obtained by scaling the normal vector found earlier by \tau\in\Bbb R; translate this line by adding the vector \langle6,4,4\rangle to get the line we want,

\vec\rho(\tau)=\langle6,4,4\rangle+\langle0,-20,15\rangle\tau

\boxed{\vec\rho(\tau)=\langle6,4-20\tau,4+15\tau\rangle}

6 0
3 years ago
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