The remaining area that the grass will cover will be given by:
Area=[area of rectangle]-[area of the triangle]
area of rectangle will be:
A1=length*width
=20*8
=160 square feet

The area of the triangle will be:
A2=1/2absin theta
since the triangle is equilateral, all the angles are equal and the size of each angle is 60°. Thus the area will be:
A2=1/2*5*5sin 60
=18.3
thus the remaining area of the grass will be:
Area=160-18.3
=149.7 square feet

6 0

3 years ago

Explain why vertical angles cannot be adjacent angles?

iris [78.8K]

Adjaadjacent angles are angles that are the vertex but vertical angles are lines that cross each other.

6 0

2 years ago

A two column table with 5 rows. The first column, x, has the entries, negative 3, negative 1, 1, 3. The second column, y, has th

Bezzdna [24]

Answer:

The correct answer is y = 3x + 1.

Step-by-step explanation:

I did the assignment, and it was correct. Hope it helps!

8 0

2 years ago

Read 2 more answers

A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. The average nicotine content from a

tatuchka [14]

Answer:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

$df=n-1=15-1=14$

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

Step-by-step explanation:

Information provided

$\bar X=42.6$ represent the average nicotine content

$s=3.7$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =40$ represent the value to check

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:

Null hypothesis: $\mu \leq 40$

Alternative hypothesis: $\mu > 40$

The statistic for this case would be:

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

And replcing we got:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

The degrees of freedom are:

$df=n-1=15-1=14$

The p value would be:

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

6 0

3 years ago