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Verdich [7]
3 years ago
11

❤Please answer❤Yesterday a parking lot had 10 cars parked in the parking lot, 40% were silver, 3 were red, and the rest were blu

e. How many of the cars were blue?
A. 2 Cars
B. 5 Cars
C. 3 Cars
D. 8 Cars
Mathematics
2 answers:
velikii [3]3 years ago
5 0
The answer is c. 3 because 40% of 10=4+3=7. 10-7=3 therefore, the answer is three
Illusion [34]3 years ago
3 0

Answer:

C: 3 Cars

Step-by-step explanation:

40% of 10 is 4.

10 - 4 ( silver cars ) = 6

6 - 3 ( red cars ) = 3

So there are 3 blue cars.

^.^

- Amanda

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If 20% of a number is 35, what is 50% of that number? 55, 70, 87.5, or 175
Dimas [21]
Here is what I did to find the answer: 
20% of n is 3520% * n = 3520/100 * n = 35n = 35 * 100 / 20n = 175
50% of 175 = 50% * 175 = 50/100 * 175 = 87.5
The correct answer would be 87.5.
4 0
3 years ago
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Show that √(1-cos A/1+cos A) =cosec A - cot A​
Gennadij [26K]

Hi there!

\sqrt{\frac{1-cosA}{1+cosA}} =

We can begin by multiplying by its conjugate:

\sqrt{\frac{1-cosA}{1+cosA}} * \sqrt{\frac{1+cosA}{1+cosA}}  = \\\\\sqrt{\frac{(1-cosA)(1 + cosA)}{(1+cosA)(1 + cosA)}} =

Simplify using the identity:

1 - cos^2A = sin^2A

\sqrt{\frac{(1-cos^2A)}{(1+cosA)^2}} =\\\\\sqrt{\frac{(sin^2A)}{(1+cosA)^2}} =

Take the square root of the expression:

{\frac{sinA}{1+cosA} =

Multiply again by the conjugate to get a SINGLE term in the denominator:

{\frac{sinA}{1+cosA} * {\frac{1-cosA}{1-cosA} =\\

Simplify:

{\frac{sinA(1-cosA)}{1-cos^2A} =

Use the above trig identity one more:

{\frac{sinA(1-cosA)}{sin^2A} =

Cancel out sinA:

{\frac{(1-cosA)}{sinA}  =

Split the fraction into two:

{\frac{1}{sinA}   - \frac{cosA}{sinA} =

Recall:

1/sinA = cscA\\\\cosA/sinA = cotA

Simplify:

\frac{1}{sinA} + \frac{cosA}{sinA} = \boxed{cscA - cotA}

5 0
2 years ago
5x + 5x =10x____^a0​
PSYCHO15rus [73]
5(x + 10)?  A.5x + 10 B.5x + 10x C.5x + 50 D.5 + x + 10.
7 0
3 years ago
High-power experimental engines are being developed by the Stevens Motor Company for use in its new sports coupe. The engineers
Maurinko [17]

Answer:

The 99% of a confidence interval for the average maximum HP for the experimental engine.

(536.46, 603.54)

Step-by-step explanation:

<u><em>Step:-1</em></u>

Given that the mean of the Population = 540HP

Given that the size of the sample  'n' = 9

Given that the mean of the sample = 570HP

Given that the sample standard deviation = 30HP

<u><em>Step(ii):-</em></u>

<u><em>Degrees of freedom = n-1 =9-1 =8</em></u>

<u><em>t₀.₀₀₅ = 3.3554</em></u>

The 99% of a confidence interval for the average maximum HP for the experimental engine.

(x^{-} - t_{\frac{0.01}{2} ,8} \frac{S.D}{\sqrt{n} } ,x^{2}  + t_{\frac{0.01}{2},8 } \frac{S.D}{\sqrt{n} } )

(570-3.354\frac{30}{\sqrt{9} } , 570+3.354\frac{30}{\sqrt{9} } )

(570 - 33.54 , 570+33.54)

(536.46 , 603.54)

<u><em>Final answer</em></u> :-

<em>The 99% of a confidence interval for the average maximum HP for the experimental engine.</em>

<em>(536.46, 603.54)</em>

<em></em>

4 0
3 years ago
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earnstyle [38]

Answer:

C is the answer ,got it right on EDG 2020

Step-by-step explanation:

8 0
3 years ago
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