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Arada [10]
2 years ago
8

Please helpppp I will give you a brainleist !!!!!! ​

Mathematics
2 answers:
Verdich [7]2 years ago
4 0

Answer:

Hope this helps

Step-by-step explanation:

4. 10^2

5. 64 / 14 * 9 + 8

64/126+8

8 64/126

8 4/9

6.   1.762 x 10^3

Lyrx [107]2 years ago
3 0

Answer:

I am confused with #5 but the rest are solid and checked

Step-by-step explanation:

#4 (A)

10*10=100

#5 (97)

64+4*4+3*3+2*2*2=

64+16+9+8=97

#6 (1.762 × 103)

#7 1070

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Savatey [412]
The answer is 1.2 b/c you have to solve x by simplifying both sides of the equation and isolating the variable.
5 0
3 years ago
A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of
Masteriza [31]

Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 15.4 inches, and standard deviation of 3.5 inches.

This means that \mu = 15.4, \sigma = 3.5

16 items are chosen at random

This means that n = 16, s = \frac{3.5}{\sqrt{16}} = 0.875

What is the probability that their mean length is less than 16.8 inches?

This is the p-value of Z when X = 16.8. So

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

By the Central Limit Theorem

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

Z = \frac{16.8 - 15.4}{0.875}

Z = 1.6

Z = 1.6 has a p-value of 0.9452.

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

5 0
2 years ago
I want to thank everyone on here who has helped me!
wariber [46]
Omg that so nice I love this app sm
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2 years ago
Read 2 more answers
0.00000000828860 in scientific notation
azamat

0.00000000828860 upto which decimal?


6 0
3 years ago
Which statement correctly describes order of operations (PEMDAS)?
Serjik [45]

Answer:

B

Step-by-step explanation:

parentheses exponents multiplication division addition subtraction

4 0
2 years ago
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