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kumpel [21]
3 years ago
7

Two lawn care companies are competing for a landscaping contract. Company A charges a $250 monthly consulting fee plus $10 per s

quare foot of landscaping, whereas Company B charges a $90 monthly consulting fee plus $12 per square foot of landscaping. For how many square feet of landscaping will the two companies charge the same amount in one month?
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
4 0

Answer:

Company A: 250+10n

Company B: 90+12n

Step-by-step explanation:

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Verify the following identity and show steps<br><br> (Cos2 θ)/(1+sin2 θ)= (cot θ-1)/(cot θ+1)
Paul [167]

Answer:

Verified below

Step-by-step explanation:

We want to show that (Cos2θ)/(1 + sin2θ) = (cot θ - 1)/(cot θ + 1)

In trigonometric identities;

Cot θ = cos θ/sin θ

Thus;

(cot θ - 1)/(cot θ + 1) gives;

((cos θ/sin θ) - 1)/((cos θ/sin θ) + 1)

Simplifying numerator and denominator gives;

((cos θ - sin θ)/sin θ)/((cos θ + sin θ)/sin θ)

This reduces to;

>> (cos θ - sin θ)/(cos θ + sin θ)

Multiply top and bottom by ((cos θ + sin θ) to get;

>> (cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ)

In trigonometric identities, we know that;

cos 2θ = (cos² θ - sin²θ)

cos²θ + sin²θ = 1

sin 2θ = 2sinθcosθ

Thus;

(cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ) gives us:

>> cos 2θ/(1 + sin 2θ)

This is equal to the left hand side.

Thus, it is verified.

5 0
2 years ago
Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a rando
DENIUS [597]

Answer:

n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =73,\sigma =10.5)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

We want to find the value of n that satisfy this condition:

P(71.5 < \bar X

And we can use the z score formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we have this:

P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z

And we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94

And by properties of the normal distribution we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z

If we solve for P(Z we got:

P(Z

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: z=-1.881

And using this we have this equality:

-1.881 = -0.14286 \sqrt{n}

If we solve for \sqrt{n} we got:

\sqrt{n} = \frac{-1.881}{-0.14286}=13.167

And then n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

6 0
3 years ago
The sale is for $30 this is 75 % of the original price find the original price
Sauron [17]
1. $52.80

2. 80

3. 46 people

4. 15 minutes

5. 2%

6. 147.03

I think these answers are correct but I am not 100% sure
6 0
4 years ago
I need helpppppppp!!!
yarga [219]

Answer: i would say probably 3 or -3

I hope it helps

3 0
3 years ago
Whats the square root of 25ab^6
r-ruslan [8.4K]
5b^3 sq rt of a

There's not enough a's
7 0
3 years ago
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