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REY [17]
2 years ago
10

the table shows the height of the sage bush at each of 2 ages. what was the bushes average rate of growth per year during this t

ime period. pls answer quick

Mathematics
1 answer:
stellarik [79]2 years ago
3 0

Answer:

2.4 inches a year

Step-by-step explanation:

It took 5 years for the bush to grow 22 inches, and it went up by 12 inches in 5 years. So, the easiest thing to do is 12 / 5 = 2.4 inches a year. :)

(Brainliest ??)

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5 divided by what equals 1.21
salantis [7]

Answer:

4.132

Step-by-step explanation:


5 0
2 years ago
In rhombus ABCD, What is m ? AEB?
Zinaida [17]
80 degree angel i believe
3 0
3 years ago
Which value is a solution to the inequality x < 4?
densk [106]
The answer is x > 4.5.

Explanation:

-24 > -6(x - 0.5)

-24 > -6x + 3

6x > 3 + 24

6x > 27

x > 27/6

x > 4.5

Hope this helps :)
3 0
2 years ago
The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.
Anton [14]

Answer:

a) The evidence suggest the true mean life of a biomedical device > 5500

b) (5487.94, +∞)

c) There is evidence to support that the mean is equal to 5500

Step-by-step explanation:

Here we have

(a) To test the hypothesis we have the claim that the mean life of biomedical devices > 5500

Therefore, we put the null Hypothesis which is the proposition that a difference does not exist. That is

H₀:  μ = 5500

Therefore, the alternative hypothesis will be

Hₐ:  μ > 5500

The test statistic is then found by;

t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}} =\frac{5617.8-5500 }{\frac{234.5 }{\sqrt{15}}} \approx 1.95

The P value from the T table at df = n - 1 = 15 - 1 = 14 is

0.025 < P < 0.05

The P value is given as 0.036 from the T distribution at 14 derees of freedom df

Therefore, since P < α or 0.05, we reject the null hypothesis. That is we fail to reject Hₐ:  μ > 5500. The evidence suggest the true mean life of a biomedical device > 5500

(b) The 95% confidence interval is given as

CI=\bar{x}\pm t_\alpha \frac{s}{\sqrt{n}}

Which gives    t_\alpha =  \pm2.145

CI=5617.8\pm 2.145 \times \frac{234.5}{\sqrt{15}}

The confidence interval of the lower bound on the mean is then

(5487.94, +∞)

c) From the above result, we find that the mean of 5500 is contained in the range of the lower confidence on the mean. We can therefore, accept the null hypothesis. That is there is evidence to support that the mean is equal to 5500.

6 0
3 years ago
Please help me with the questions
GREYUIT [131]

Answer:

\cos(B)  =  \frac{12}{15}

Step-by-step explanation:

\cos(B)  =  \frac{adjacent}{hypotenuse}

\cos(B)  =  \frac{12}{15}

6 0
2 years ago
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