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Nata [24]
3 years ago
5

In a greenhouse, basil is grown in planters that have a base of 4 1 4 in by 1 2 3 in. If there is a 4 × 4 array of these planter

s with no space between them, what is the area covered by the planters?
Mathematics
1 answer:
Kryger [21]3 years ago
8 0

Answer:

The area covered by the planters is 113.6 square inches.

Step-by-step explanation:

Assuming that the dimmensions of the greenhouse are 4 1/4 by 1 2/3 inches. We first need to convert them from mixed fractions to normal fractions, we do that by summing the integer part to the fraction one. We have:

length = 4 + 1/4 = 16/4 + 1/4 = 17/4 inches

width = 1 + 2/3 = 3/3 + 2/3 = 5/3 inches

The area of the greenhouse is given by:

area = length*width

area = (17/4)*(5/3) = 85/12 = 7.1 square inches

Since there is an array of 4x4 there is a total of 16 greenhouses and it's total area is 16*7.1 = 113.6 square inches

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2 years ago
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Answer:

θ ≈ 71.6°

Step-by-step explanation:

The angle between two lines with slopes m₁ and m₂ is:

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Here, m₁ = -2 and m₂ = 1.

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3 0
4 years ago
A number has exactly three factors and is less than 50 what is this number
vfiekz [6]
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3 years ago
A function is given: f (x) = 2x + 3<br> What is the value of f(-2)?
Nadusha1986 [10]

Answer:

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A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i
garri49 [273]

Answer:

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df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

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And for this case the rejection region would be:

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Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :  

Null hypothesis:\mu \leq 5  

Alternative hypothesis:\mu > 5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0
3 years ago
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