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lana66690 [7]
3 years ago
12

PLEASE HELP, MATH, PLEASESolve for n: n+5/16 = -1​

Mathematics
1 answer:
Anit [1.1K]3 years ago
3 0

\tt Step-by-step~explanation:

To solve for n, we have to isolate n. To do so, we move all the terms that are not n to one side of the equation, and leave n on the other side.

\tt Steps:

Equation: n + 5/16 = -1

Subtract 5/16 on both sides to bring it to the right side of the equation.

\tt n+5/16-5/16=-1-5/16\\\\n=-\frac{21}{16}~or~-1\frac{5}{16}

\Large\boxed{\tt Our~final~answer:~n=-\frac{21}{16}~or~-1\frac{5}{16} }

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A barge moves 7km/hr in still water. It travels 45km upriver and 45km downriver in a total time of 14 hr. What is the speed of t
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Answer:

Step-by-step explanation:

2(3x−5)+4x−10=4x+2(x+6)

Step 1: Simplify both sides of the equation.

2(3x−5)+4x−10=4x+2(x+6)

(2)(3x)+(2)(−5)+4x+−10=4x+(2)(x)+(2)(6)(Distribute)

6x+−10+4x+−10=4x+2x+12

(6x+4x)+(−10+−10)=(4x+2x)+(12)(Combine Like Terms)

10x+−20=6x+12

10x−20=6x+12

Step 2: Subtract 6x from both sides.

10x−20−6x=6x+12−6x

4x−20=12

Step 3: Add 20 to both sides.

4x−20+20=12+20

4x=32

Step 4: Divide both sides by 4.

4x/4 =32/4

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3 years ago
Brad reads his favorite book for 1 1/4 hours each day. If he has read the book for 22 1/2 hours so far, how many days has he rea
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Answer:

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Tay–Sachs Disease Tay–Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carri
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Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

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Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability that their offspring will develop the disease is approximately .25.

This means that p = 0.25

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This means that n = 3

Question a:

This is P(X = 3). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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Question b:

This is P(X = 1). So

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Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

p = 0.75*0.75*0.25 = 0.1406

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

6 0
3 years ago
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Hey there!

Let's first define our goal. Our goal here is to be able to isolate the variable, n, and solve for it by moving everything to the other side. We see that we have two sides of the equation, and by using the property of equality which states that whatever you do to one side you do to the other and the expression remains equal, we can solve this equation.

It looks like the only thing we have on the left side with n is that 6 1/2. In order to get rid of it, we must subtract it from both sides, because the property of equality states you must do the inverse operation.

We have:

n = 12 - 6 1/2

n = 5 1/2

Hope this helps!
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