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gtnhenbr [62]
3 years ago
5

Sea level is considered to be 0 feet. A probe released at sea level dropped at a constant rate for 3.25 minutes, reaching an ele

vation of −35.75 feet relative to sea level.
What was the probe's elevation relative to sea level after the first minute?
Mathematics
1 answer:
mash [69]3 years ago
8 0
Thank you for posting your question here at brainly. The probe's elevation relative to sea level after the first minute, the probe dropped 11 feet per minute, therefore, it dropped 11 feet in the first minute. I hope the answer helps you. 
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Evaluate the following.<br> (iii) sin 30° + tan 40° - cosec 60° / cot 45° + cos 60° - sec 30°​
bearhunter [10]

Sin 30 = 1/2

tan 45 = 1

Cosec 60 = 2 / √3 = 2√3 / 3

cot 45 = 1

Cos 60 = 1/2

sec 30 = 2 / √3 = 2√3 / 3

_______________________________

[ 1/2 + 1 - 2√3/3 ] ÷ [ 1 + 1/2 - 2√3/3 ] = <em>1</em>

The face and denominator of the fraction are exactly the same thus the answer is 1 .

4 0
3 years ago
Una heladeria dispone de 20 frutas distintas para elaborar sus malteadas. Si los clientes pueden elegir tres sabores para mezcla
Dahasolnce [82]

Answer:

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

Step-by-step explanation:

En este caso, el cliente que adquiere una malteada de tres sabores distintos sigue el siguiente procedimiento:

1) El primer sabor sale de cualquiera de las 20 frutas disponibles.

2) El segundo sabor es distinto al primer sabor, es decir, que sale de las 19 frutas restantes.

3) El tercer sabor es distinto al primer sabor y al segundo sabor, es decir, que sale de las 18 frutas restantes.

Puesto que existe una doble conjunción y que puede importar el orden según la preferencia del cliente, se habla matemáticamente de una permutación, definida como:

n\mathbb{P}k = \frac{n!}{(n-k)!} (1)

Donde:

n - Número de sabores disponibles, adimensional.

k - Número de sabores escogidos, adimensional.

Si tenemos que n = 20 y k = 3, entonces la cantidad de malteadas de tres sabores distintos es:

n\mathbb{P}k = \frac{20!}{(20-3)!}

n\mathbb{P}k = \frac{20!}{17!}

n\mathbb{P}k = 20\cdot 19\cdot 18

n\mathbb{P}k = 6840

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

5 0
3 years ago
Mrs. Computer is 3 times older than her daughter, Mousy. The sum of their ages is 52. How old is Mousy?
grandymaker [24]
Mousy=m
Mrs. Computer=3m
m+3m=52
4m=52
m=52/4
m=13
mousy=13, Mrs. Computer=3*13=39 
8 0
3 years ago
Use the following function rule to find f(28).
IrinaVladis [17]

What following function tool?

7 0
3 years ago
Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five event
mezya [45]

Answer:

The probability of Javier volunteers for less than three a month is 25%

Step-by-step explanation:

According to the questions,

The probability of Javier to attend exactly 5 events :-

P ( x = 5 ) = 25% = \frac{1}{4}

The probability of Javier to attend exactly 4 events :-

P ( x = 4 ) = \frac{1}{4}

The probability of Javier to attend exactly 3 events :-

P ( x = 3 ) = \frac{1}{4}

The probability of Javier to attend exactly 2 events :-

P ( x = 2 ) = \frac{3}{20}

The probability of Javier to attend exactly 1 events :-

P ( x = 1 ) = \frac{1}{20}

The probability of Javier to attend no events :-

P ( x = 0 ) = \frac{1}{20}

Now we have to calculate the probability of Javier volunteers for less than three P ( x < 3 ).

P ( x < 3 ) = P ( x = 0 ) + P ( x = 1 ) + P ( x = 2 )

P ( x < 3 ) = \frac{1}{20} + \frac{1}{20} + \frac{3}{20}   = \frac{1 + 1 + 3}{20}  = \frac{5}{20}  = \frac{1}{4}   = 0.25 = 25 %.

8 0
3 years ago
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