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Sholpan [36]
3 years ago
12

Divide: 407:10 Guys I need help I don’t know how to do divide SOMEONE HELP ME QUICKLY!!!

Mathematics
2 answers:
natulia [17]3 years ago
8 0

Answer:

407/10= 40.7

Step-by-step explanation:

katen-ka-za [31]3 years ago
7 0

Answer:

40

7 Remainder

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FREE BRAINLIEST. PLEASE ANSWER STEPBY STEP 1. 2. 3.
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Answer:

8

Step-by-step explanation:

First do whats in the parenthesis:

(8+18/9)

Using PEMDAS again, we do the division first:

(8+2)

Now addition:

(10)

Now subtraction:

(10)-2

8

3 0
3 years ago
Hiii please help i’ll give brainliest if you give a correct please please thank youuuuu
Finger [1]

Answer: 24 boxes of pens

Step-by-step explanation:

4 0
3 years ago
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Taylor bought her favorite teacher, Mrs. Miranda, a new tissue box holder. Regular sized tissue boxes are 9 inches by 4 inches.
Artist 52 [7]

Answer:

9x4=36

Step-by-step explanation:

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3 years ago
What line is parallel to y=1/2x+5
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7 0
3 years ago
A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard d
bagirrra123 [75]

The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>

Suppose that we have:

  • Sample size n > 30
  • Sample mean = \overline{x}
  • Sample standard deviation = s
  • Population standard deviation = \sigma
  • Level of significance = \alpha

Then the confidence interval is obtained as

  • Case 1: Population standard deviation is known

\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}

  • Case 2: Population standard deviation is unknown.

\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}

For this case, we're given that:

  • Sample size n = 90 > 30
  • Sample mean = \overline{x} = 138
  • Sample standard deviation = s = 34
  • Level of significance = \alpha = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: Z_{0.1/2} = ±1.645

Thus, we get:

CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

Learn more about confidence interval for population mean from large samples here:

brainly.com/question/13770164

3 0
2 years ago
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