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Volgvan
3 years ago
8

How to estimate the sum of using different methods. 187,176,154,207

Mathematics
1 answer:
nadya68 [22]3 years ago
7 0

Answer:

tuff

guess you will fail

Step-by-step explanation:

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Your chemistry textbook has 936 pages and weighs 4.2 pounds. your calculus textbook has 1368 pages and weighs 6.0 pounds. these
Zarrin [17]
Let the number of chemistry books be x and the number of calculus books be y
total number of books in each case is:
x+y=24
x=24-y.....i

Weight of each case:
2707.2/20
=135.36
thus
4.2x+6.0y=135.36..ii substituting i in ii
4.2(24-y)+6y=135.36
100.8-4.2y+6y=135.36
hence
1.8y=34.56
y=19.2
hence
x=24-19.2
x=4.8
thus the number of chemistry books was 4.8*20=96 number of calculus was
19.2*20
=384 books

4 0
3 years ago
Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight
GrogVix [38]

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

simplifying the equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the  95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$.

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

substitute the values in the above equation, we get

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\

simplifying the above equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

brainly.com/question/14825274

#SPJ4

6 0
2 years ago
What is the value of x
LenKa [72]

Answer: x= 98°

Step-by-step explanation:

Concept:

Here, we need to know the concept of the "Exterior Angle theorem"

The exterior angle theorem states that the measure of an exterior angle is equal to the sum of the measures of the two opposite angles of the triangle.

If you are still confused, you may refer to the attachment below for a graphical explanation.

Solve:

x = 53 + 45 = 98°

Hope this helps!! :)

Please let me know if you have any questions

7 0
3 years ago
How do these numbers compare? Drag the correct comparison symbol to the box. 6.32 6.320
sweet [91]

Answer:

6.32 = 6.320

Step-by-step explanation:

Convert both to like decimal by adding 0 to the end. Then compare.

6.320   =  6.320

8 0
3 years ago
Red 25 green 10 blue 15 probability of the green
Leto [7]
Total = 25 + 10 + 15 = 50
Green = 25

P(green) = 25/50  = 1/2

Answer: 1/2
7 0
4 years ago
Read 2 more answers
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