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mr Goodwill [35]
3 years ago
10

A bee flies 6 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for 11 minutes, and then fli

es directly back to the hive at 11 feet per second. It is away from the hive for a total of 13 minutes. A. what equation can you use to find the distance of the flowerbed from the hive? B. How far is the flowerbed from the hive? Let d be the distance of the flowerbed from the hive.
Mathematics
1 answer:
zhenek [66]3 years ago
4 0

Did you ever get the answer

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All six printing machines together can print 222 pages in a minute how many pages can each machine print per hour
Furkat [3]
The answer should be 13,320
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Two brothers went shopping at a back to school sale where all shirts and shorts were the same price. The younger brother spent $
salantis [7]

Answer:

Each shirt costs 10$ cause there is a 10 $ difference between the two brothers and the only thing the older brother didn't buy was another shirt.

Have a great day. Brainliest always helps!!!

8 0
3 years ago
Are snow days a thing of the past or do they still have a place in today's world? Answer whether or not you believe the Mayor an
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5 0
3 years ago
(7x-y=7<br>(x+2y=6 estimate the solution to the system of equations ​
jolli1 [7]

Multiply the first equation by 2 to then use elimination by adding the 2 equations.

So, 14x-2y=14

+. x+2y=6

Equals 15x=20 so x=20/15=4/3

Plug in to first equation to get y

7(4/3)-y=7

28/3-y=7

28/3-y=21/3

So y=7/3

Plug both x and y into second equation to check

4/3+2(7/3)=

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So x=4/3, y=7/3

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8 0
4 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

6 0
3 years ago
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