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2 years ago

Which of the following equations have infinitely many solutions?

Mathematics

1 answer:

ikadub [295]2 years ago

4 0

The equation which represents a system with infinitely many solutions is;

<h3>What system of equations have infinitely many solutions as in the task content?</h3>

The condition for a situation in which case an equation has infinitely many solutions is such that the right hand side and left hand side of the equation are equal.

On this note, it follows that the answer choices which represents the equations with infinitely many solutions is;

46x+23=46x+23

Read more on infinitely many solutions;

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pav-90 [236]

Answer:

hope it's helpful

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7 0

2 years ago

Please help me! Quickly I WILL GIVE YOU BRAINLEST IF YOU ANSWER ME PLEASE! 15points!

hram777 [196]

Answer:

You would plot the following points, remembering that they’re in the form (x,y) = (time, height):

(1,35) (2,60) (3,92) (4,120) (5,175)

The ordered pairs would be closer together, and there would be 4 times more of them if you were to measure every 15 seconds instead. You would have a line with more points that were closer together.

5 0

3 years ago

According to a study conducted by the Gallup Organization, the the proportion of Americans who are afraid to fly is 0.10. A rand

notsponge [240]

Answer: 0.1457

Step-by-step explanation:

Let p be the population proportion.

Given: The proportion of Americans who are afraid to fly is 0.10.

i.e. p= 0.10

Sample size : n= 1100

Sample proportion of Americans who are afraid to fly = $\hat{p}=\dfrac{121}{1100}=0.11$

We assume that the population is normally distributed

Now, the probability that the sample proportion is more than 0.11:

$P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]$

Hence, the probability that the sample proportion is more than 0.11 = 0.1457

3 0

3 years ago

What is 4/9 • t when t=1/3

Reptile [31]

$\frac{4}{9}* \frac{1}{3}= \frac{4}{27}$

8 0

3 years ago

Read 2 more answers

Which situation will have the greatest number of permutations?

Anton [14]

Answer:

Option C will have greatest number of permutations.

Step-by-step explanation:

The number of license plates with 4 letters followed by 3 numbers, where letters and digits cannot be repeated.

We have total 26 letters (A_Z) and the letters can't be repeated. So we have 26 letter options for 1st letter

25 letters options for 2nd letter

24 letter options for 3rd letter

23 letter options for 4th letter.

So total options for 4 letters 26*25*24*23 = 358800

Now, 3 numbers are required. We have total 10 numbers (0-9)

10 number options for 1st number

9 number options for 2nd number

8 number options for 3rd number

Total options for 3 numbers= 10*9*8 = 720

total options for 4 letters and 3 numbers will be: 358800 * 720 = 258,336,000

The number of license plates with 3 letters followed by 4 numbers, where letters and digits cannot be repeated.

We have total 26 letters (A_Z) and the letters can't be repeated. So we have 26 letter options for 1st letter

25 letters options for 2nd letter

24 letter options for 3rd letter

So total options for 3 letters= 26*25*24 = 15600

Now, 4 numbers are required. We have total 10 numbers (0-9)

10 number options for 1st number

9 number options for 2nd number

8 number options for 3rd number

7 number options for 4th number

Total options for 4 numbers = 10*9*8*7 = 5040

total options for 4 letters and 3 numbers will be: 15600 * 5040 = 78,624,000

C. The number of license plates with 4 letters followed by 3 numbers, where letters and digits can be repeated.

Now, the letters and digits can be repeated. So we have 26 options for each 4 letters and 10 options for each 3 numbers

26*26*26*26 *10*10*10 = 456,976 * 10,00 = 456,976,000

total options for 4 letters and 3 numbers will be: 456,976,000

The number of license plates with 3 letters followed by 4 numbers, where letters and digits can be repeated.

Now, the letters and digits can be repeated. So we have 26 options for each 3 letters and 10 options for each 4 numbers

26*26*26 *10*10*10*10 = 17,576 * 10,000 = 175,760,000

total options for 3 letters and 4 numbers will be: 175,760,000

So, Option C will have greatest number of permutations.

7 0

3 years ago