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3 years ago

10

36 POINTS! Write a problem where the negative square root is not an integer. then graph the square root(tell me what to graph si

nce we cant on here)

1 answer:

Tcecarenko [31]3 years ago

7 0

<span>" - √5 "....and 'square root ' is statement and difficult to graph....you need a function</span>

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Which is a solution of x2 – x – StartFraction 3 Over 4 EndFraction = 0?

Anna [14]

Answer:

x = - $\frac{1}{2}$ , x = $\frac{3}{2}$

Step-by-step explanation:

Given

x² - x - $\frac{3}{4}$ = 0 ( add $\frac{3}{4}$ to both sides )

x² - x = $\frac{3}{4}$

Solve using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2( - $\frac{1}{2}$ )x + $\frac{1}{4}$ = $\frac{3}{4}$ + $\frac{1}{4}$

(x - $\frac{1}{2}$ )² = 1 ( take the square root of both sides )

x - $\frac{1}{2}$ = ± 1 ( add $\frac{1}{2}$ to both sides )

x = $\frac{1}{2}$ ± 1

Thus

x = $\frac{1}{2}$ - 1 = - $\frac{1}{2}$

x = $\frac{1}{2}$ + 1 = $\frac{3}{2}$

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3 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

3 years ago