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3 years ago

The graph of a system of equations with the same slope will have no solutions. (1 point)

1 answer:

3 0

I don’t really get what you’re asking here but a system of 2 parallel lines has no solution, and a system of 2 parallel lines is called inconsistent.

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HELP PLS!

Zigmanuir [339]

Answer:

x = 2 is the correct answer :)

5 0

2 years ago

Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.

lesya [120]

Answer:

0.1994 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851$

P(sample of 20 boxers is more than 167 pounds)

$P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)$

$= 1 - P(z \leq 0.8438)$

Calculation the value from standard normal z table, we have,

$P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%$

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0

2 years ago

How did you do? Rate your work on a scale of 1 to 5, with 5 as the highest score. Then write a brief evaluation of your work bel

Arte-miy333 [17]

Answer:

there have answer use solutions problem or write in note

Step-by-step explanation:

maybe if I can answer

6 0

3 years ago

Read 2 more answers

Please try your best and show your work and I will give brainliest!

dimulka [17.4K]

Answer:

Part A: impossible

Part B: Either equal or blue

Part C: 9 green and 2 blue were added

Step-by-step explanation:

Part A:

The only colors included in this problem are red, blue, and green. There is no black colored pencil, therefore, it is impossible to get one from the box.

Part B:

I'm not sure what you're asking in this question, but I will give you the two choices. If it is before the additional 11 colored pencils are added to the box, the chance of drawing a red and the chance of drawing a blue will be equal, because both of them have 11 of each color. If it is after the additional 11 colored pencils are added to the box, then the chance of drawing a blue colored pencil will be greater than the chance of drawing a red colored pencil. After the 11 colored pencils are added, there are 13 blue and 11 red. The blue is greater.

Part C:

The least number of green colored pencils added has to be 9, because the chance of drawing a green pencil is now greater than the chance of drawing a red pencil. If we add 8 more green pencils, the likelihood would be the same. Therefore, the number of green colored pencils added has to be at least 9. If we have the last 2 colored pencils be blue, then there would be 11 red, 13 blue, and 12 green. This fits all the conditions, therefore, adding 9 green colored pencils and 2 blue colored pencils is the answer.

I hope this helps and please mark me as brainliest!

5 0

2 years ago

The number sentence 4x6=6x4 is an example of what property

Nana76 [90]

The commutative property of multiplication

4 0

2 years ago