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velikii [3]
3 years ago
11

Each month Adam changes the amount of food and sunlight a potted plant receives. When he started the plant was 63 cm tall. In th

e first month it grew 14 cm. The following month the plant grew 3 times as much as it did the first month. How many meters tall is the plant at the end of the second month?
Mathematics
2 answers:
iren [92.7K]3 years ago
4 0
At the end of the second month, the plant is 1.19 meters tall.
Umnica [9.8K]3 years ago
4 0
The plant is 119 m tall at the end of the second month.
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The area of a rectangle is expressed as (12m+18) square feet. If the width of the rectangle is 6 feet, choose an expression to r
Naddik [55]

a = l \times w \\ (12m + 18) = l \times 6 \\  \frac{12m + 18}{6}  = l \\ (2 m + 3) = l

the answer is C.

4 0
3 years ago
A study reported that the average age of menarche of girls entering the first year class of a small American private college in
Verizon [17]

Answer:

Null hypothesis: \mu = 11.89

Alternative hypothesis: \mu \neq 11.89

And we want to test this with a confidence interval:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

After conduct the interval of 95% for the true mean we got: (11.59, 13.31)

And for this case the confidence interval contains the value of 11.89, so then we have enough evidence to FAIL to reject the null hypothesis at 5% of significance that the true mean is 11.89.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=12.98 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=30 represent the sample size  

Solution to the problem

For this case we want to test the following hypothesis:

Null hypothesis: \mu = 11.89

Alternative hypothesis: \mu \neq 11.89

And we want to test this with a confidence interval:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

After conduct the interval of 95% for the true mean we got: (11.59, 13.31)

And for this case the confidence interval contains the value of 11.89, so then we have enough evidence to FAIL to reject the null hypothesis at 5% of significance that the true mean is 11.89.

4 0
3 years ago
Divide 6⁄13 by 6⁄12 . <br><br> A. 1⁄12<br> B. 9⁄16<br> C. 12⁄13<br> D. 13⁄12
nadezda [96]

6/13÷6/12=12/13 so C.

Hope this helps!

5 0
2 years ago
Read 2 more answers
In the straightedge and compass construction of the equilateral triangle below, how do you know that AB ≈ AC?
Ostrovityanka [42]

Answer:

C

Step-by-step explanation:

C:  AB is the radius of both circles.

5 0
3 years ago
Read 2 more answers
QUESTION 1 1 POINT<br> Factor: 27m3<br> 64.<br> Provide your answer below:<br> Content attribution
pickupchik [31]

Answer:

\boxed{ \bold{ \boxed{ \sf{(3m - 4)(9 {m}^{2}  + 12m + 16)}}}}

Step-by-step explanation:

\sf{27 {m}^{3}  - 64}

\dashrightarrow{ \sf{( {3m)}^{3}  -  {(4)}^{3} }}

Use the formula of a³ - b³ = ( a - b) ( a² + ab + b² )

\longrightarrow{ \sf{(3m - 4)(3 {m}^{2}  + 3m \times 4 +  {4}^{2} }})

\longrightarrow{ \sf{(3m - 4)( {9m}^{2}  + 12m + 16}})

Hope I helped!

Best regards! :D

5 0
3 years ago
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