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3 years ago

Express b+1/3b-2 with “b” as the subject

Mathematics

1 answer:

lara [203]3 years ago

4 0

Answer:

b = $\frac{1+2a}{3a-1}$

Step-by-step explanation:

Given

a = $\frac{b+1}{3b-2}$ ( multiply both sides by 3b - 2 )

a(3b - 2) = b + 1 ← distribute left side

3ab - 2a = b + 1 ( subtract b from both sides )

3ab - b - 2a = 1 ( add 2a to both sides )

3ab - b = 1 + 2a ← factor out b from each term on the left side

b(3a - 1) = 1 + 2a ( divide both sides by 3a - 1 )

b = $\frac{1+2a}{3a-1}$

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Jam world amusement park sells special multi-day passes for guest who want to visit the park for more than a day there most popu

zalisa [80]

Answer:

The 3-day pass costs $16.67 more per day than the 7-day pass.

Step-by-step explanation:

The easiest way to do this is to find the cost of each pass for the same amount of days. To do this, we find the lowest common multiple of 3 and 7 which would be 21

Lowest common multiple = $7*3 = 21$

Next we find the cost of each pass for 21 days by multiplying cost of the 7 day pass by 3, and the cost of the 3 day pass by 7

Cost of 7 day pass for 21 days = $364*3 = 1092$

Cost of 3 day pass for 21 days = $206*7=1442$

From this, we know that the difference between the two passes is $350 per 21 days. Since the question asks how much more would the 3-day pass cost per a day than the 7-day pass, we have to divide the difference by 21

$350/21 = 16.66667$

We can round this to $16.67. As a result, the 3-day pass costs $16.67 more per day than the 7-day pass.

3 0

3 years ago

Read 2 more answers

Evaluate 5-(2w-8)=6w-9

Brilliant_brown [7]

Step-by-step explanation:

Steps

5−(2w−8)=6w−9

Show Steps

Expand 5−(2w−8): −2w+13

−2w+13=6w−9

Subtract 13 from both sides

−2w+13−13=6w−9−13

Simplify

−2w=6w−22

Subtract 6w from both sides

−2w−6w=6w−22−6w

Simplify

−8w=−22

Divide both sides by −8

−8w

−8 =

−22

−8

Show Steps

Simplify

click her

5 0

3 years ago

Of all the registered automobiles in a city, 12% fail the emissions test. Fourteen automobiles are selected at random to undergo

postnew [5]

Answer:

a) 0.1542

b) 0.7685

c) 0.2315

d) No, it is not unusual

Explanation:

The procedure to make the test meets the requirements of binomial experiments because:

there are two possible mutually exclusive outputs: fail the test, or pass the test.
the probability of each event remains constant during all the test (p=12% = 0.12, for failing the test, and 1-p = 88% = 0.88, for passing the test)
each trial (test) is independent of other trial.

Solution

(a) Find the probability that exactly three of them fail the test.

You want P(X=3)

Using the equation for discrete binomial experiments, the probability of exactly x successes is:

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{(n-x)}$

Substituting C(n,x) with its developed form, that is:

$P(X=x)=\dfrac{n!}{x!\cdot (n-x)!}\cdot p^x\cdot (1-p)^{(n-x)}$

Thus, you must use:

x = 3 (number of automobiles that fail the emissions test)
n = 14 (the number of automobiles selected to undergo the emissions test),
p = 0.12 (probability of failing the test; this is the success of the variable on our binomial experiment)
1 - p = 0.88 (probability of passing the test; this is the fail of the variable on our binomial experiment)

$P(X=3)=\dfrac{14!}{3!\cdot (14-3)!}\cdot 0.12^3\cdot 0.88^{11}=0.1542$

(b) Find the probability that fewer than three of them fail the test.

The probability that fewer than three of them fail the test is the probability that exactly 0, or exactly 1, or exactly 2 fail the test.

That is: P(X=0) + P(X=1) + P(X=2)

Using the same formula:

$P(X=0)=\dfrac{14!}{0!\cdot 14!}\times 0.12^0\cdot 0.88^{14}$

$P(X=0)=0.1670$

$P(X=1)=\dfrac{14!}{1!\cdot 13!}\cdot 0.12^1\cdot 0.88^{13}$

$P(X=1)=0.3188$

$P(X=2)=\dfrac{14!}{2!\cdot 12!}\codt0.12^2\cdot 0.88^{12}$

$P(X=2)=0.2826$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.7685

(c) Find the probability that more than two of them fail the test.

The probability that more than two of them fail the test is equal to 1 less the probability that exactly 0, or exactly 1, or exactly 2 fail the test:

P( X > 2) = 1 - P( X = 0) - P(X = 1) - P(X = 2)

P X > 2) = 1 - [P(X=0) + P(X = 1) + P(X = 2)]

P (X > 2) = 1 - [0.7685]

P (X > 2) = 0.2315

(d) Would it be unusual for none of them to fail the test?

Remember that not failing the test is the fail of the binomial distribution. Thus, none of them failing the test is the same as all of them passing the test.

You can find the probability that all the automibles pass the emission tests by multiplying the probability of passing the test (0.88) 14 times.

Then, the probability that none of them to fail the test is equal to:

$(1-p)^{14}\\\\(0.88)^{14}=0.1671$