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s344n2d4d5 [400]
2 years ago
12

Plzzzzzz help. I have to waist so many points becouse peaple dont answer my question plzzzzzz help will give brainliest (pic dow

n below)
A model house that runs completely on renewable energy uses 222 windmills and 333 solar panels. Each windmill generates MMM kilowatt-hours (kWh) of energy per year, and each solar panel generates SSS kWh per year.
Which expressions can we use to describe how many kWh of energy the model house will output in 101010 years?
Choose 2 answers:

Mathematics
1 answer:
Ronch [10]2 years ago
5 0

Answer:

10(2M+3S)

20M+30S

Step-by-step explanation:

we know that

The number of kWh of energy that the model house generate per year is given by the sum of  kWh of energy generates by the windmills multiplied by the number of windmills, plus the  kWh of energy generates by the solar panels multiplied by the number of solar panels.

so

2M+3S

For 10 year

Multiply the expression above by 10

10(2M+3S)

Apply distributive property

20M+30S

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Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
2 years ago
Read 2 more answers
Pens cost 15 pence each.
ch4aika [34]

Answer:

The amount the school pays is £32.40

Step-by-step explanation:

The cost of each pen = 15 pence

The cost of each ruler = 20 pence

The number of pens bought by the school = 150

The number of rulers bought by the school = 90

The cost reduction (discount) on the items bought = 1/5

Therefore, we have;

The total cost of the pens bought by the school = 150 × 15 = 2250 = £22.50

The total cost of the rulers bought by the school = 90 × 20 = 1800 = £18.00

The total cost of the writing materials (rulers and pens) bought by the school = £22.50 + £18.00 = £40.50

The discount = 1/5 total cost reduction = 1/5×£40.50  = $8.10

The amount the school pays = The total cost of the writing materials - The discount

The amount the school pays = £40.50 - $8.10 =  £32.40

The amount the school pays =  £32.40.

7 0
1 year ago
Suppose bug A lands on the end of the blade farthest from the pivot. Assume the wiper turns through an angle of 120°. In one cyc
ruslelena [56]

Answer:

s = 21.33*π in

Step-by-step explanation:

Given:

- The complete data is given in the figure (attached):

Solution:

- The bugs lands on the end of the wiper blade furthest to the pivot. The arc length (s) swept by a wiper in one cycle i.e θ = 120° would be given by:

                                    s = 2*r*θ

- Where, r: The distance between the bug and the pivot = 16 in  

              θ : Angle swept in radians

- The distance travelled by the bug is s:  

                                   s = 2*(16)*( 120 / 180 ) * π  

                                   s = 21.33*π in

3 0
1 year ago
Vineet solved a system of equations by substitution. In his work, he substituted an expression for one of the variables and solv
mezya [45]
Vineet can conclude that:
B ) the system of equations does not have a solution.
10 0
2 years ago
Read 2 more answers
Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in
Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                         P.Q.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of Americans who decide to not go to college = 48%

           n = sample of American adults = 331

           p = population proportion of Americans who decide to not go to

                 college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                        of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.645) = 0.90

P( -1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < \hat p-p < 1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

P( \hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

<u>90% confidence interval for p</u> = [ \hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } } , 0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } } ]

 = [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error =  Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }

              0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }

              \sqrt{n}  = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}

              \sqrt{n} = 54.79

               n = 54.79^{2}

               n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0
2 years ago
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