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Agata [3.3K]
3 years ago
15

Find the value of y.

Mathematics
2 answers:
Jobisdone [24]3 years ago
6 0

Answer:

3

Step-by-step explanation:

LUCKY_DIMON [66]3 years ago
4 0
The answer would be 3
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Two of the cards of an ordinary deck of 52 cards are lost. What is the probability that a random card drawn from this (defective
Nostrana [21]

Answer:

0.72

Step-by-step explanation:

Given :Two of the cards of an ordinary deck of 52 cards are lost.

To Find : What is the probability that a random card drawn from this (defective) deck is a spade?

Solution:

Total number of cards = 52

2 cards are lost

Remaining cards = 50

When no spade card is lost

So, No. of spade cards = 13

Total cards = 50

So, probability of getting spade = \frac{13}{50}

When 1 spade card is lost

So, No. of spade cards = 12

Total cards = 50

So, probability of getting spade = \frac{12}{50}

When 2 spade cards are lost

So, No. of spade cards = 11

Total cards = 50

So, probability of getting spade = \frac{11}{50}

So, The he probability that a random card drawn from this (defective) deck is a spade =  \frac{13}{50}+\frac{12}{50}+\frac{11}{50}

                  =  \frac{36}{50}

                  =  [tex]0.72/tex]

Hence the probability that a random card drawn from this (defective) deck is a spade is 0.72

3 0
3 years ago
Multiple choice question
Studentka2010 [4]
The last one I can't get the signs up
3 0
4 years ago
Read 2 more answers
HELP PLEASE NO FOOLISH ANSWERS
JulsSmile [24]

Answer:

Mass determines how fast a car can accelerate. Since a car accelerates slower with more adults, and faster with less adults, this means that the mass has a big impact on acceleration. Therefore, we can conclude that the mass determines how fast a car can accelerate.

3 0
3 years ago
The equation 13x + 1 = 5 – x can be solved by graphing y = 13x + 1 and y = 5 – x.
shusha [124]

Answer:

The graph shows that when x =3 both

\frac{1}{3}x + 1

and 5 – x equal 2

Step-by-step explanation:

The equation 13x + 1 = 5 – x can be solved by graphing y = 13x + 1 and y = 5 – x.

From the graph, the two lines representing the two equations intersect at (3,2).

This implies that the solution to the system

y =  \frac{1}{3} x + 1

and

y=5-x

is x=3, y=2

Hence the value of the two functions equals 2 at x=3.

7 0
3 years ago
As the domain values approach infinity, the range values approach infinity. As the domain values approach negative infinity, the
Firdavs [7]

Limits at infinity truly are not so difficult once you've become familiarized with then, but at first, they may seem somewhat obscure. The basic premise of limits at infinity is that many functions approach a specific y-value as their independent variable becomes increasingly large or small. We're going to look at a few different functions as their independent variable approaches infinity, so start a new worksheet called 04-Limits at Infinity, then recreate the following graph.

plot(1/(x-3), x, -100, 100, randomize=False, plot_points=10001) \ .show(xmin=-10, xmax=10, ymin=-10, ymax=10) Toggle Explanation Toggle Line Numbers

In this graph, it is fairly easy to see that as x becomes increasingly large or increasingly small, the y-value of f(x) becomes very close to zero, though it never truly does equal zero. When a function's curve suggests an invisible line at a certain y-value (such as at y=0 in this graph), it is said to have a horizontal asymptote at that y-value. We can use limits to describe the behavior of the horizontal asymptote in this graph, as:

 and 

Try setting xmin as -100 and xmax as 100, and you will see that f(x) becomes very close to zero indeed when x is very large or very small. Which is what you should expect, since one divided by a large number will naturally produce a small result.

The concept of one-sided limits can be applied to the vertical asymptote in this example, since one can see that as x approaches 3 from the left, the function approaches negative infinity, and that as x approaches 3 from the right, the function approaches positive infinity, or:

 and 

Unfortunately, the behavior of functions as x approaches positive or negative infinity is not always so easy to describe. If ever you run into a case where you can't discern a function's behavior at infinity--whether a graph isn't available or isn't very clear--imagining what sort of values would be produced when ten-thousand or one-hundred thousand is substituted for x will normally give you a good indication of what the function does as x approaches infinity.

6 0
4 years ago
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