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Nikolay [14]
1 year ago
9

Use the difference between -2 and 5 to find the distance. The distance is

Mathematics
2 answers:
Juli2301 [7.4K]1 year ago
8 0

Answer:

-2 - 5 = -7

Step-by-step explanation:

pantera1 [17]1 year ago
8 0

Answer:

It must be probably <u>√</u><u>2</u><u>9</u>

Step-by-step explanation:

distance =   \sqrt{  { (- 4)}^{2} +  {(5)}^{2} }

hope it to be the correct...

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The zeros of function d are -3 and 8. Which expression could be a factor of function d?
marysya [2.9K]
X+3 or x-8
These answers will both work!
8 0
3 years ago
How do i write 2m in a verbal expression?
eduard
2m: "two multiplied by m"
3 0
3 years ago
Read 2 more answers
20x^2-30=4x solve for x
ankoles [38]

Answer:

x=1.3288 or -1.1288

Step-by-step explanation:

Given 20x^{2}-30=4x

    Subtract 4x from both sides

         20x^{2}-4x-30=0

Since we can factor out 2 from whole equation, let's factor out 2.

         2(10x^{2}-2x-15)=0

Divide with 2 on both sides

         10x^{2}-2x-15=0

We got the quadratic equation, we can solve for x using quadratic formula.

x=\frac{-b\pm\sqrt{b^{2}-4ac} } {2a}

x=\frac{2\pm\sqrt{2^{2}-4X10X(-15)} } {2X10}

x=\frac{2\pm\sqrt{604} }{20}

x=\frac{2\pm2\sqrt{151}}{20}

x=1.3288or-1.1288


8 0
3 years ago
Read 2 more answers
Each bag of apples weighs 4 1/2 lb how much do 6 bags weigh?
Inessa [10]

Answer:

27 pounds

Step-by-step explanation:

You need to multiply 4.5 by 6.

Hope this helps.

7 0
2 years ago
A recent study found that the average length of caterpillars was 2.8 centimeters with a
pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

\mu = 2.8, \sigma = 0.7.

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{4 - 2.8}{0.7}

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0
2 years ago
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