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2 years ago

Can you evaluate (g ○ f)(0)? Explain why or why not.

Mathematics

2 answers:

chubhunter [2.5K]2 years ago

6 0

Answer:

No you can't because just like multiplication, you cant divide anything by zero.

Step-by-step explanation:

You must evaluate the function f first. Division by 0 is undefined.

Therefore, The composition cannot be evaluated.

Snowcat [4.5K]2 years ago

4 0

Answer:

sample answer on edg2020

Step-by-step explanation:

To evaluate the composition, you need to find the value of function f first. But, f(0) is 1 over 0, and division by 0 is undefined. Therefore, you cannot find the value of the composition.

You might be interested in

Which statements accurately describe the function f(x) = 3(StartRoot 18 EndRoot) Superscript x? Select three options. The domain

Studentka2010 [4]

Answer:

The domain is all real numbers.

The initial value is 3

The simplified base is $9\sqrt{2}$

Step-by-step explanation:

The given function is $f(x)=3(\sqrt{18})^x$ .

To find the initial value, we put x=0 into the function:

$f(0)=3(\sqrt{18})^0$ .

$f(0)=3(1)=3$ .

The initial value is actually 3.

The given function is an exponential function, therefore the domain is all real numbers.

The range of this function refers to all values of y for which the function is defined.

The line y=0, is the horizontal asymptote.

The range is $y\:>\:0$

The simplified base is $3\sqrt{18}=3\sqrt{9\times2}$ .

$3\sqrt{18}=3\sqrt{9}\times\sqrt{2}$

$3\sqrt{18}=3\times3\times\sqrt{2}$

$3\sqrt{18}=9\sqrt{2}$

4 0

2 years ago

Read 2 more answers

An artist wants to make alabaster pyramids using a block of alabaster with a volume of 576 cubic inches. She plans to make each

tensa zangetsu [6.8K]

Given:
volume of the block = 576 cubic inches
Square pyramid = base area : 3 square inches ; height = 4 inches

Volume of square pyramid = a² * h/3
V = 3 square inches * 4inches/3
V = 4 cubic inches

volume of block / volume of square pyramid
576 cubic inches / 4 cubic inches = 144

She can make 144 square pyramids.

3 0

2 years ago

Read 2 more answers

AC ≅ AE; ∠ACD ≅ ∠AED

pshichka [43]

Answers:

A) △ACF ≅ △AEB because of ASA.

D) ∠CFA ≅ ∠EBA

E) FC ≅ BE

Solution:

AC ≅ AE; ∠ACD ≅ ∠AED Given

The angle ∠CAF ≅ ∠EAB, because is the same angle in Vertex A

Then △ACF ≅ △AEB because of ASA (Angle Side Angle): They have a congruent side (AC ≅ AE) and the two adjacent angles to this side are congruent too (∠ACD ≅ ∠AED and ∠CAF ≅ ∠EAB), then option A) is true: △ACF ≅ △AEB because of ASA.

If the two triangles are congruent, the ∠CFA ≅ ∠EBA; and FC ≅ BE, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), then Options D) ∠CFA ≅ ∠EBA and E) FC ≅ BE are true

4 0

2 years ago

Read 2 more answers

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

Ugo [173]

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:

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:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

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

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

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.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, 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

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

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

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

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

This probability, using continuity correction, 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.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

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

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

Using continuity correction, 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

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

8 0

2 years ago

You jog 6/23 miles around a track each day. If you jogged that distance 4 times last week, how many miles did you jog?

LenaWriter [7]

Well, since you jogged 6/23 mi. a day, for 4 days, it'll be (6/23)×4. This is 24/23 which is one mile and 1/24 of one.

6 0

2 years ago