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icang [17]
2 years ago
14

On a coordinate plane, a line connects (negative 4, negative 5) and (negative 2, negative 4), a curve connects (negative 2, nega

tive 4) and (0, 0), a curve connects (0, 0) and (2, 4), and a line connects (2, 4) and (4, 5). The graph shows a function f(x). Determine the following values. f –1(4) = f –1(–5) = f –1(0) =
Mathematics
1 answer:
statuscvo [17]2 years ago
6 0

Answer: f^-1(4)= 2 f^-1(-5)= -4 f^-1(0)= 0

Step-by-step explanation:

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The answer is 2.9. You would round up since there is a 5

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8. Carisa is four inches taller than her friend Lily.
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Factor the expression completely -15t-6
serg [7]

Answer:

\large\boxed{-15t-6=-3(5t+2)}

Step-by-step explanation:

-15t-6=(-3)(5t)+(-3)(2)=-3(5t+2)

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3 years ago
Y Varies directly as X and inversely as Z. Y=100 when X=5 and Z=10 Find Y when X=3 and Z=60
Law Incorporation [45]

Answer:

y = 10

Step-by-step explanation:

based on the question if y  varries directly as x

mathematically

y ∝ x

also, y varries inversly as z can be mathematically expressed as

y ∝ 1/z

combining the two expressions

y ∝ x ∝ 1/z

i. e

y = kx/z..... where k is the constant of proportionality

make  k the subject of formulae

yz = kx

Divide both sides by x

k =yz/x

when y=100 , x = 5 z =10

k = 100 × 10/5

k = 200

to find y when x = 3 and z = 60

<h3>from the equation connecting x,y,z</h3>

k =yz/x

200 =60y/3

cross multiply

60y = 200 × 3

60y = 600

divide both sides by 60

y = 10

5 0
2 years ago
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 12
melamori03 [73]

Answer:

98.75% probability that every passenger who shows up can take the flight

Step-by-step explanation:

For each passenger who show up, there are only two possible outcomes. Either they can take the flight, or they do not. The probability of a passenger taking the flight is independent from other passenger. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

The probability that a passenger does not show up is 0.10:

This means that the probability of showing up is 1-0.1 = 0.9. So p = 0.9

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets

This means that n = 125

Using the approximation:

\mu = E(X) = np = 125*0.9 = 112.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.9*0.1} = 3.354

(a) What is the probability that every passenger who shows up can take the flight

This is P(X \leq 120), so this is the pvalue of Z when X = 120.

Z = \frac{X - \mu}{\sigma}

Z = \frac{120 - 112.5}{3.354}

Z = 2.24

Z = 2.24 has a pvalue of 0.9875

98.75% probability that every passenger who shows up can take the flight

7 0
2 years ago
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