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2 years ago

What is 2+3 and how do you solve that answer

Mathematics

2 answers:

LenKa [72]2 years ago

4 0

Answer: 5

Work: I just did mental math, but the proper way to do it is to get a number line, and put a dot at 2. Since you need to move three spots ahead, you move the dot three times to the right, and it will land on 5.

<em>I hope this helps, and Happy Holidays! :)</em>

Elden [556K]2 years ago

3 0

5. Put 2 of fingers up . Then put 3 more fingers up . Count the total

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The first quartile of a data set is 23, the median is 30,the third quartile is 33, and an outlier is 50. Which of these data val

Len [333]

Answer:50

Step-by-step explanation:

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2 years ago

In your computation, approximate its value as 3.14.

andrey2020 [161]

Answer: d

Step-by-step explanation:

Find the area of the rectangle.

$A=l*w\\A=(15in)(12in)\\A=180in^2$

Find the area of the base of the semicircle.

$A=\pi r^2$

12 in would be the diameter, divide by 2 to get the radius.

$\frac{12}{2}=6$

Plug this into the formula.

$A=(3.14)(6)^2\\A=113.04in^2$

Since we do not have a full circle but a semi one instead, we must divide our result by 2.

$\frac{113.04}{2}=56.52in^2$

Add both areas.

180+56.52=236.52 $in^2$

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2 years ago

irga5000 [103]

152*3=456............,

7 0

2 years ago

12. A man planted trees on both sides of

netineya [11]

Answer:

1000 trees : Can u plz give me the brainliest answer? Plz!!!!!!!!!!!!!!!

Step-by-step explanation:

5000m divided by 5m = 1000 trees

Have a nice day!!!!!! :-)

3 0

2 years ago

Read 2 more answers

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

2 years ago