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andrew-mc [135]
3 years ago
7

What is the range of y=−x2−6x+1?

Mathematics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

All real numbers / (-∞, ∞) / R

Step-by-step explanation:

The equasion provided is a linear equasion meaning it can be written in the y=ax+b format, where both a and b are real numbers. All linear eqasions have a range of all real numbers, you can write this using the three solutions provided.

You can solve this two ways, number one would be to put this in a graphing calculator and number two would be reducing the equasion to the y=ax+b format.

<h2>Method 1:</h2>

look at image provided. The app is PhotoMath.

<h2>Method 2:</h2>

Write the equasion:

y = -2x - 6x + 1

Collect like terms:

y = -8x + 1

This now proves that the equasion is linear becuase you can write it in the y=ax+b format. a being -8 and b being 1.

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Hey there!!!

Remember :

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ratelena [41]

Option C ,

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levacccp [35]

Answer:

<h2><em>m</em><em><</em><em>1</em><em>=</em><em>6</em><em>2</em><em>°</em></h2>

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<em>or</em><em>,</em><em> </em><em><</em><em>1</em><em>+</em><em><</em><em>1</em><em>+</em><em>5</em><em>6</em><em>=</em><em>1</em><em>8</em><em>0</em><em>°</em>

<em>or</em><em>,</em><em> </em><em>2</em><em><</em><em>1</em><em>=</em><em>1</em><em>8</em><em>0</em><em>°</em><em>-</em><em>5</em><em>6</em><em>°</em>

<em>or</em><em>,</em><em> </em><em>2</em><em><</em><em>1</em><em>=</em><em>1</em><em>2</em><em>4</em>

<em>or</em><em>,</em><em> </em><em><</em><em>1</em><em>=</em><em>1</em><em>2</em><em>4</em><em>/</em><em>2</em>

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6 0
4 years ago
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zheka24 [161]
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4 0
3 years ago
Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the fiv
Dahasolnce [82]

Answer and Explanation:

Given : Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029

\sum P(X)=1

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

\mu=\sum xP(x)=2.5

The standard deviation of the random sample is

Variance=\sigma^2

\sigma^2=\sum x^2P(x)-\mu^2

\sigma^2=7.436-(2.5)^2

\sigma^2=7.436-6.25

\sigma^2=1.186

\sigma=\sqrt{1.186}

\sigma=1.08

Therefore, The mean is 2.5 and the standard deviation is 1.08.

5 0
3 years ago
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