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4 years ago

Evaluate n/6+2 when n=12

Mathematics

2 answers:

garik1379 [7]4 years ago

6 0

Answer:

Step-by-step explanation:

nikklg [1K]4 years ago

4 0

Answer:

$\frac{3}{2} \: or \: 1.5$

Step-by-step explanation:

$\frac{12}{6 + 2} = \frac{12}{6 + 2} = \frac{12}{8} = \frac{6}{4} = \ \frac{3}{2} \: or \: 1.5$

You might be interested in

4y+3=3y+x in slope intercept form

nirvana33 [79]

Answer:

y = x - 3

Step-by-step explanation:

4y + 3 = 3y + x

y + 3 = x

y = x - 3

4 0

3 years ago

A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received wit

m_a_m_a [10]

Answer:

The probability that the message will be wrong when decoded is 0.05792

Step-by-step explanation:

Consider the provided information.

To reduce the chance or error, we transmit 00000 instead of 0 and 11111 instead of 1.

We have 5 bits, message will be corrupt if at least 3 bits are incorrect for the same block.

The digit transmitted is incorrectly received with probability p = 0.2

The probability of receiving a digit correctly is q = 1 - 0.2 = 0.8

We want the probability that the message will be wrong when decoded.

This can be written as:

$P(X\geq3) =P(X=3)+P(X=4)+P(X=5)\\P(X\geq3) =\frac{5!}{3!2!}(0.2)^3(0.8)^{2}+\frac{5!}{4!1!}(0.2)^4(0.8)^{1}+\frac{5!}{5!}(0.2)^5(0.8)^0\\P(X\geq3) =0.05792$

Hence, the probability that the message will be wrong when decoded is 0.05792

4 0

3 years ago

I need help please and thank you

marta [7]

So what u need to do is

46+x=180 and u will get the answer

5 0

3 years ago

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of

belka [17]

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==> t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348

5 0

3 years ago

The situation is as follows: Smart phones have changed the world and how we spend our time. A group of researchers at BYU want t

Scilla [17]

Answer:

The 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019 is between 306.65 minutes and 317.35 minutes.

Step-by-step explanation:

Confidence interval normal

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.96}{2} = 0.02$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.02 = 0.98$ , so Z = 2.054.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.054\frac{54}{\sqrt{430}} = 5.35$

The lower end of the interval is the sample mean subtracted by M. So it is 312 - 5.35 = 306.65 minutes

The upper end of the interval is the sample mean added to M. So it is 312 + 5.35 = 317.35 minutes

The 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019 is between 306.65 minutes and 317.35 minutes.

8 0

3 years ago