Which two transformations were preformed

What is the probability of not drawing a card 6 or greater from a regular deck of cards?

Mamont248 [21]

Ok so there are 52 cards in a deck if you take all the cards that have the number 6 which is only 4 cards so 52-4 = 48 cards without any 6 so the ratio is 48/52
chances.

If you're trying to take any card that is greater than 6 its a little more complex there is T,2,3,4,5,6,7,8,9,10,B,Q,K so you have to tack away 6,7,8,9,10,B,Q,K from 4 rows so 8 x 4 =32 cards 52-32= 20 so the probability of not having any cards greater than 6 is 20/52

3 0

3 years ago

HELP QUICK <br> (question in picture)

anzhelika [568]

Answer:

Step-by-step explanation:

There’s a total of 10+16+20+18 = 64 candies.

There’s 10 red candies.

10 / 64 = 0.15625

Rounded: 0.2

7 0

3 years ago

A deposit of $35.45 was made to a checking account. Before the deposit, the account had a balance of −$12.39 .

Naddika [18.5K]

The answer would be 23.06.

7 0

3 years ago

Find the measure of ∠FEG.<br> A) 25° <br> B) 65° <br> C) 80° <br> D) 90°

MissTica

First, you have to set 5x equal to (3x+10). They are vertical angles and therefore congruent.

5x = 3x+10 The variable should only be on one side of the equation,

5x - 3x = 10 Therefore subtract 3x from both sides

2x = 10 Divide by 2 on both sides.

x = 5

We're not done yet, since we've found the x-value we have to solve for ∠GEC.

∠GEC = 3x+10 = 3(5)+10 = 15+10 = 25°

∠GEC and ∠FEG form a right angle. This means that the sum of the two angles is 90°

90° = 25° + x In this case, we're using "x" to solve for ∠FEG

90-25 = x We subtract 25 from both sides to get the variable by itself.

65° = x

<h2>∠FEG = 65° The Answer is B</h2>

6 0

3 years ago

Read 2 more answers

NO LINKS!!! What is the equation of these two graphs?

S_A_V [24]

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

Step-by-step explanation:

<h3><u>Question 1</u></h3>

<u>Method 1 - modelling as a quadratic with restricted domain</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

$y=ax^2+bx+c$

Given points:

(-4, 2)
(-1, 4)
(4, 6)

Substitute the given points into the equation to create 3 equations:

<u>Equation 1 (-4, 2)</u>

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

<u>Equation 2 (-1, 4)</u>

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

<u>Equation 3 (4, 6)</u>

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

<u>Method 2 - modelling as a square root function</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, and the x-intercept should be included, we can model this curve as a <u>square root function</u>.

Given points: