Convert 7/8 to decimal

How do you do this ??????????????????????

kicyunya [14]

Answer:

b

Step-by-step explanation:

you can take the rectagel and you x the rectagel

6 0

3 years ago

Lines s and t are perpendicular if the slope of line s is 5 what is the slope of line t

ycow [4]

The slope of line T is -1/5, because perpendicular lines have opposite reciprocal slopes, meaning the opposite sign (positive or negative) and flipped (whole number becomes a fraction).

5 0

3 years ago

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Which is equivalent sin^-1 (cos(pi/2))?Give your answer in radians

NNADVOKAT [17]

Answer:

Given the expression: $\sin^{-1}(\cos(\frac{\pi}{2}))$

Let the value of the given expression in radians be $\theta$

then;

$\sin^{-1}(\cos(\frac{\pi}{2})) =\theta$

$\cos \frac{\pi}{2} = \sin \theta$ ......[1]

We know the value of $\cos \frac{\pi}{2} = 0$

Substitute the given value in [1] we have;

$\sin \theta = 0$

Since, the value of $\sin \theta$ is 0, therefore, the value of $\theta$ is in the form of:

$\theta = n\pi$ ; where n is the integer.

At n =0, 1 and 2, {Since, n is the integer}

Value of $\theta =0, \pi$ and $2\pi$

therefore, the answer in radians either $0 , \pi$ or $2\pi$

6 0

3 years ago

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Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case

olga nikolaevna [1]

Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> = Mike will show up.

Given:

$P(D^{c})=0.55\\P(M^{c})=0.45$

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

$P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25$

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

$=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75$

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

$P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25$

Thus, the probability that neither Dave nor Mike will show up is 0.25.

6 0

3 years ago

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Because of a virus introduced into a bacterial culture, the population of the bacteria decreasedfrom 3000 to 1000 in 50 minutes.

Mamont248 [21]

Answer:

208

Step-by-step explanation:

Given that;

y= a(1 - b)^t

Where;

wherein y is the final amount, a is the original amount, b is the decay factor, and x is the amount of time that has passed.

To obtain the constant b,

y= ae^-bt

Substituting values;

1000 = 3000 e^-b(50)

1000/3000 = e^b(50)

0.33 = e^b-(50)

ln (0.33) = ln(e^b(-50))

-1.1 = -50b

b= 1.1/50

b= 0.022

After two hours or 120 minutes;

y= 3000(1 - 0.022)^120

y= 207.8661772

y= 208

8 0

3 years ago