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hjlf
3 years ago
7

Find the square root of 2.2 correct to two place of decimal​

Mathematics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

1.48

Step-by-step explanation:

my calculator said sqrt(2.2)=1.48

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What is 10 +10×2÷8+5​
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Answer:

just use a calculator next time 17.5

Step-by-step explanation:

its 17.5 my dude

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The math counts club had a party at school.there were 20 cookies and 40 slices of pizza to be shared equally.each student had th
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there would be 4 students

5 cookies for each

10 slices of pizza for each

hope this helps

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The product of two numbers subtracted by 44, the result is 100. Find the numbers.​
melamori03 [73]

Answer: the numbers can be 16 and 9, 6 and 24 or other factors of 144

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4 years ago
Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.
zysi [14]

Answer:

a) 78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

b) 78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d)   ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

Step-by-step explanation:

Part a

For this case we have the following data given

\bar X = 78.25 represent the sample mean for the customer order totals

\sigma =22.50 represent the population deviation

n= 40 represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be \alpha=0.1 and \alpha/2 = 0.05 and the critical value from the normal standard distirbution would be given by:

z_{\alpha/2}=1.64

And the confidence interval is given by:

\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

And replacing we got:

78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

 ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

3 0
3 years ago
How do you round 0.0695712 to 2 significant figures?
BartSMP [9]

Answer:

0.07

Step-by-step explanation:

.069 is close to 0.070 and since it only asks for two so you only put 0.07

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