All categories

3 years ago

What happens when y = -f(-x)?

Mathematics

1 answer:

IgorLugansk [536]3 years ago

3 0

Answer:

It's a reflection over the x axis.

Step-by-step explanation:

Idk I googled it.

You might be interested in

Which expression is equivalent to (m8n-4) 1/4

ser-zykov [4K]

Answer:

D. $\dfrac{m^2}{n}$

Step-by-step explanation:

I assume this is your problem:

$(m^8n^{-4})^{\frac{1}{4}}$

If so, then this is the solution.

$(m^8n^{-4})^{\frac{1}{4}} =$

$= m^{8 \times \frac{1}{4}} n^{-4 \times \frac{1}{4}}$

$= m^2 n^{-1}$

$= \dfrac{m^2}{n}$

4 0

3 years ago

Read 2 more answers

1/4kg of flour with 2/9kg of suger how much all togather

mote1985 [20]

Answer:

17/36

Step-by-step explanation:

8 0

3 years ago

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

erastovalidia [21]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Using the normal approximation to the binomial.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

175 visitors, so $n = 175$

46.7% through the Beaver Meadows, so $p = 0.467$

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

This is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

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

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

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

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

6.3% over the Grand Lake park entrance, so $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability is P(X < 12 - 0.5) = P(X < 11.5), which is the pvalue of Z when X = 11.5. So

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

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

22.7% with no recorded point, so $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

This probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

99.78% probability that more than 55 visitors have no recorded point of entry

3 0

3 years ago

Find the hcf and the lcm of these for 10 points plz <br>

neonofarm [45]

Answer:

LCM of 64 and 94= 3008

HCF of 64 and 94= 2

Step-by-step explanation:

Do a venn diagram with the prime factors of 64 on the left, and the prime factors of 94 on the right. If they have the same factors put them in the middle of the venn diagram. Then you multiply everything in the diagram and get the LCM. The HCF is found by multiplying the factors in both lists which is only 2.

4 0

3 years ago

Correct me if I'm wrong, isn't the Distance from A to C be 7?

Alex_Xolod [135]

Answer:

no you're not wrong go with your gut

Step-by-step explanation:

5 0

3 years ago