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3 years ago

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 angle6t,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