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Rudik [331]
3 years ago
9

In a deck of cards with 52 cards and 4 suits, what is the probability of getting an ace?

Mathematics
2 answers:
DedPeter [7]3 years ago
7 0

The probability of picking up an ace in a 52 card deck, where there are 4 aces is 4/54 (simplify to 2/27).

Katarina [22]3 years ago
6 0

The probability of getting a ace is 8/52

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Jason had $32. He spent all the money on buying 4 CDs for $x each and 2 magazines for $y each. If Jason had bought 5 CDs and 2 m
Leni [432]
You can set up two equations from the information given. I will set them up for you:

32 = 4x + 2y
36 = 5x + 2y

Let's solve the first equation to come up with a value for y.

32 = 4x + 2y
32 - 4x = 2y
16 - 2x = y

Now we plug y into the other equation.

36 = 5x + 2(16-2x)
36 = 5x + 32 - 4x
4 = x

Now we have our real x value and we can plug it into the first equation.

32 = 4(4) + 2y
32 = 16 + 2y
16 = 2y
8 = y

Since x = 4 and y = 8, you get the final coordinates of (4,8).

Your answer is the second statement provided above.
3 0
3 years ago
Pint 16oz $3.98
LiRa [457]
Okay, so when dealing with unit rates you want to see how much you have of your main measurement. In this case your main measurement is ounces. Since you have 16 and 32 you are going to divide your dollar amount by those and round to the nearest tenth. 
Here are the equations that need to be used. Figure out the answers to them and round. This will give you your answer
3.98/16=
5.98/32=
4 0
3 years ago
How many odd numbers are there between twenty and forty?
inna [77]

Answer:

Total 10

Step-by-step explanation:

21. 23, 25, 27, 29, 31 , 33, 35, 37, 39.

hope this helps you

5 0
2 years ago
Identify the interval on which the quadratic function is positive.
Alenkasestr [34]

Answer:

\textsf{1. \quad Solution:  $1 < x < 4$,\quad  Interval notation:  $(1, 4)$}

\textsf{2. \quad Solution:  $-2 < x < 4$,\quad  Interval notation:  $(-2, 4)$}

Step-by-step explanation:

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

The intervals on which a <u>quadratic function</u> is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.  

As the given <u>quadratic function</u> has a negative leading coefficient, the parabola opens downwards.   Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}

Apply the <u>zero-product property</u>:

\implies x-1=0 \implies x=1

\implies x-4=0 \implies x=4

Therefore, the interval on which the function is positive is:

  • Solution:  1 < x < 4
  • Interval notation:  (1, 4)

<h3><u>Question 2</u></h3>

The intervals on which a <u>quadratic function</u> is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.  

As the given <u>quadratic function</u> has a positive leading coefficient, the parabola opens upwards.   Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}

Apply the <u>zero-product property</u>:

\implies x+2=0 \implies x=-2

\implies x-4=0 \implies x=4

Therefore, the interval on which the function is negative is:

  • Solution:  -2 < x < 4
  • Interval notation:  (-2, 4)
3 0
1 year ago
Brianna has 3 posters she wants to hang on the wall. How many different ways can
Sedbober [7]

Answer:

print them out or something

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