All categories

3 years ago

Please please help idk what to do

Mathematics

2 answers:

KIM [24]3 years ago

4 0

2 is the answer, it goes up 2 and over 1. Keep in mind rise over run

dimaraw [331]3 years ago

4 0

2 is the correct answer

You might be interested in

Two types of tickets were sold for an upcoming concert: adult tickets and child tickets. In total, there

Alona [7]

Answer:

Adult ticket is $60

Child ticket is $32

Step-by-step explanation:

Let x = price of adult ticket

y = price of child ticket

(1) 175x + 103y = 13796 (2) 2x + 2y = 184

x + y = 92

y = 92 - x

175x + 103(92 - x) = 13796

175x + 9476 - 103x = 13796

72x = 4320

x = 60 y = 92 - 60

y = 32

4 0

2 years ago

Dustin had $52 he got $49 more and then he spent some money dustin has $35 left how much money did dustin spend

sveta [45]

The answer: Dustin spent $66

$52+$49=$101
$101-$35=$66

5 0

3 years ago

Read 2 more answers

If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No

Effectus [21]

If A and B are independent, then $P(A\cap B)=P(A)P(B)$ .

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$

$P(A\cup B)=0.5+0.2-0.5\cdot0.2$

$\boxed{P(A\cup B)=0.6}$

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

$P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)$

$P(A^c\cap B^c)=1-0.6$

$\boxed{P(A^c\cap B^c)=0.4}$

c. DeMorgan's law can be used here too:

$P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)$

$P(A^c\cup B^c)=1-0.5\cdot0.2$

$\boxed{P(A^c\cup B^c)=0.9}$

4 0

3 years ago

From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th

wel

Answer:

$\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}$

Step-by-step explanation:

The width of a square is its side length.

The width of a circle is its diameter.

Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.

Formulas

$\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}$

$\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

$\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}$

If the diameter is equal to the side length of the square, then:
$\implies \sf r=\dfrac{1}{2}s$

Therefore:

$\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}$