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3 years ago

What is the standard deviation of the data set? 42,36,42,30,54,18

Mathematics

2 answers:

Oliga [24]3 years ago

5 0

Answer:

the standard deviations 12.2 if I'm correct.

Liono4ka [1.6K]3 years ago

3 0

the answer is 12.25!! hope this helps.

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Help me please ASAP !!!!!!!

kodGreya [7K]

Answer:

its A

PLS MARK AS BRAINLESS I MEAN BRAINLIEST :)

Step-by-step explanation:

6 0

2 years ago

-⅔ (6x +9) = 18 I really need some help and understanding with this one. Thanks in advance!!

Sever21 [200]

-⅔ (6x + 9) = 18

Apply distributive property. Multipy -⅔ to 6x and 9.

-4x - 6 = 18

Add 6 to both sides so -4x can be alone.

-4x - 6 + 6 = 18 + 6

-4x = 24

To get x alone, divide both sides by -4.

-4/-4x = 24/-4

x = -6

x = -6 is your answer. I hope this helps! Let me know if you need more help or if I got something wrong.

6 0

2 years ago

A conservative estimate of the number of stars in the universe is 6 x 10^22. The average human can see about 3,000 stars at nigh

Ilia_Sergeevich [38]

$2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

<h3><u>Solution:</u></h3>

Given that, conservative estimate of the number of stars in the universe is $6 \times 10^{22}$

The average human can see about 3,000 stars at night with only their eyes

To find: Number of times more stars are there in the universe, compared to the stars a human can see

Let "x" be the number of times more stars are there in the universe, compared to the stars a human can see

Then from given statement,

$\text{Stars in universe} = x \times \text{ number of stars human can see}$

<em><u>Substituting given values we get,</u></em>

$6 \times 10^{22} = x \times 3000\\\\x = \frac{6 \times 10^{22}}{3000}\\\\x = \frac{6 \times 10^{22}}{3 \times 10^3}\\\\x = 2 \times 10^{22-3}\\\\x = 2 \times 10^{19}$

Thus $2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

5 0

3 years ago

B. it is the area of a circle with a<br> diameter of 1. <br> True of false

Dominik [7]

It has to be true because you would need to divide

4 0

2 years ago

A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the

forsale [732]

Answer:

option (c) n = 201

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 5.5 ounce

Confidence level = 99%

Length of confidence interval = 2 ounces

Therefore,

margin of error, E = (Length of confidence interval ) ÷ 2

= 2 ÷ 2

= 1 ounce

Now,

E = $\frac{zs}{\sqrt n}$

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 = $\frac{2.58\times5.5}{\sqrt n}$

n = (2.58 × 5.5)²

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

5 0

3 years ago