Evaluating a function graphically? How do I solve this

Mathematics

1 answer:

Artyom0805 [142]3 years ago

6 0

Answer:

f(1) = -6

Explanation:

When you look at the points with x = 1, you will see a point that is open (o) and closed (•) The point that is closed (the one you are looking for) is an actual point of the function. The open point is where the function is discontinuous.

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Suppose that the sitting back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 24.4 in.

solong [7]

Answer:

$P(X \geq 26.6) = 0.0336$ , which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 24.4, \sigma = 1.2$

In this problem, a value x is significantly high if:

$P(X \geq x) = 0.01$

Using these criteria, is a back-to-knee length of 26.6 in. significantly high?

We have to find the probability of the length being 26.6 in or more, which is 1 subtracted by the pvalue of Z when X = 26.6. So

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

$Z = \frac{26.6 - 24.4}{1.2}$

$Z = 1.83$

$Z = 1.83$ has a pvalue of 0.9664.

1 - 0.9664 = 0.0336

$P(X \geq 26.6) = 0.0336$ , which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

6 0

3 years ago

Best definition combining like terms

makkiz [27]

<span>A letter that holds the place for some unknown value in mathematics VARIABLE A process where you must look for terms that have identical variable parts and then combine their coefficients. COMBINING LIKE TERMS</span>

7 0

3 years ago

Plz help..................refresh page before u answer

romanna [79]

B..

Hope this helps=D

4 0

3 years ago

Read 2 more answers

Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

NNADVOKAT [17]

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$