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3 years ago

Which of these numbers is not rational?

Mathematics

2 answers:

Veronika [31]3 years ago

7 0

Answer:

The correct option is A) $\sqrt{3}$

Step-by-step explanation:

Consider the provided numbers.

$\sqrt{3}$ , 0.25, $\frac{1}{5}$ and $\sqrt{9}$

We need to find which of them is not rational number.

Rational number: A number is said to be rational, if it is in the form of p/q. Where p and q are integer and denominator is not equal to 0. The decimal expansion of rational numbers may terminate or become periodic.

Now convert each of the number in decimal form as shown,

$\sqrt{3}=1.7320...$ ,

Here the decimal expansion of the number is neither terminating nor repeating so it is not a rational number.

0.25,

The decimal expansion of the number is terminating, so it is a rational number.

$\frac{1}{5}=0.2$

The decimal expansion of the number is terminating, so it is a rational number.

$\sqrt{9}=3$

3 is a rational number.

Hence, the correct option is A) $\sqrt{3}$

velikii [3]3 years ago

6 0

The square root of 3 which is the first choice

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What is the horizontal asymptote of the function f(x)= -2x/x+1

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<u>Answer:</u>

-2

<u>Step-by-step explanation:</u>

We have been given a function f(x)=\frac{-2x}{x+1} and we are asked to find the horizontal asymptote of our given function.

Recalling the rules for a horizontal asymptote:

1. If the numerator and denominator have equal degree, the horizontal asymptote will be the ratio of the leading coefficients.

2. If the polynomial of denominator has larger degree than the numerator, then the horizontal asymptote will be the x-axis or y=0.

3. If the polynomial of numerator has larger degree than denominator, then the function has no horizontal asymptote.

Here, the numerator and denominator are of the same degree. So the horizontal asymptote will be the ratio of the coefficients.

Horizontal asymptote = $-\frac{2}{1}$ = -2

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3 years ago

In a survey 7/8 of the people surveyed said that recycling is important. Only 1/2 of them buy products made with recycled materi

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Answer:

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3 years ago

You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of

Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}$

$P_{1}$ is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is $\frac{30}{50} = \frac{3}{5}$ .

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is $\frac{33}{53}$

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is $\frac{36}{56}$

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is $\frac{39}{59}$ . So

$P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588$

$P_{2}$ is the probability that we go R - R - B - R

$P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788$

$P_{3}$ is the probability that we go R - B - R - R

$P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076$

$P_{4}$ is the probability that we go R - B - B - R

$P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511$

$P_{5}$ is the probability that we go B - R - R - R

$P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733$

$P_{6}$ is the probability that we go B - R - B - R

$P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493$

$P_{7}$ is the probability that we go B - B - R - R

$P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476$

$P_{8}$ is the probability that we go B - B - B - R

$P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419$

So, the probability that this last marble is red is:

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768$

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that we go R-R-R-R. It is the same $P_{1}$ from the previous item(the last marble being red). So $P_{1} = 0.1588$

$P_{2}$ is the probability that we go B-B-B-B. It is almost the same as $P_{8}$ in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

$P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490$

$P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078$

There is a 20.78% probability that we actually drew the same marble all four times

3 0

3 years ago

The measure of one of the angles formed by two parallel lines and a transversal is 45°. Is it possible for the measure of any of

liubo4ka [24]

Answer: No

Step-by-step explanation:

Three of the other angles are 45°. The other four are 180° - 45°, which is not 55°. At the intersection of two lines, opposite angles are equal and adjacent angles are complementary, so two adjacent angles add to 180°. Adding a parallel line gives four more angles identical to the first four.

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4 years ago

A high school basketball coach is selecting his team. The minimum and maximum height requirements are as follows:

daser333 [38]

Answer:

B. About 2% of the boys are eligible to be a small forward on the team

Step-by-step explanation:

Recall : 1 feets = 12 inches

Point guard = 6’2" – 6’6" tall = 74 - 78 inches

Mean = 70 ; Standard deviation = 4

Z = (x - mean) / standard

P(x < 74) = (74 - 70) / 4 = 1

P(x < 78) = (78 - 70) / 4 = 2

0.97725 - 0.84134 = 0.13591

Small forward : 6'6" = 78 inches

P(x ≥ 78) = (78 - 70) / 4 = 2

P(z ≥ 2) = 0.02275 = 2.275% about 2%

Centre : 6'8" = 80

P(x ≥ 80) = (80 - 70) / 4 = 2.5

P(z ≥ 2.5) = 0.0062097 = 0.62%

8 0

3 years ago