All categories

3 years ago

Help!!!! Will give brainliest!!!!

Mathematics

1 answer:

Gwar [14]3 years ago

5 0

Answer:

is true your answer hhhhhhhhgghdhxhhchdjfjf

You might be interested in

Hannah used to make 4 bags in one day. Now she can make 7 bags in day day. What is the percent increase in the number of bags sh

pashok25 [27]

Answer:

10%

Step-by-step explanation:

???? do it look like I know oh igghht its just ahbjomd chill

7 0

3 years ago

Read 2 more answers

According to the line plot, what is the total weight of the dog toys that weighed

leva [86]

Answer:

Your que. isn't very clear. Should there be a graph or diagram? Please confirm

4 0

3 years ago

She spent 1/4 of her money on a shirt and 2/5 on shoes. What fraction of money has been spent

likoan [24]

1/4 = 5/20
2/5 = 8/20

8/20 + 5/20 = 13/20 has been spent

4 0

3 years ago

Please help fast ill give brainest

tia_tia [17]

make a graph and put the weeks and the change wich is -20 a day until the end of the four weeks

:every week 4 the city gets -20 degrees celsius colder. This means that every 4 weeks the tempature decreases then goes bac kto its origonal

8 0

3 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

3 years ago