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3 years ago

9

can you plz help me with question 9 asp and if you get the answer plz show me how you got it or i will loose points on my homewo

rk

2 answers:

marishachu [46]3 years ago

7 0

The answer is 126 feet square

olga2289 [7]3 years ago

6 0

The answer is 123 I got the answer by multiplying length ×width×hight

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X + 3= -x -5<br><br>what's x?

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-4

Step-by-step explanation:

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3 years ago

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-1/2x - 3 =4 <br> which is the value of x?

garik1379 [7]

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x = -14

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Topic: Area of non shaded area.

Dimas [21]

Answer:B) 150

Step-by-step explanation:

Area of rec= L×W

= 10×30= 300

Area of shaded area= 1/2 Base× Height

1/2×10×15=75

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3 years ago

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What is the common ratio between successive terms in a sequence?<br> 1.5, 1.2,0.96, 0.768, ...

Vanyuwa [196]

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Answer:

0.8

Step-by-step explanation:

The ratio of the first two terms is ...

1.2/1.5 = 4/5 = 0.8

That is also the ratio of successive adjacent terms.

0.96/1.2 = 0.768/0.96 = 0.8

The common ratio is 0.8.

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3 years ago

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:

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

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3 years ago