Which function represents a reflection of f(x)=3/8

Mathematics

1 answer:

Leni [432]3 years ago

8 0

Answer:

Step-by-step explanation:

Reflecting a function across the Y axis means replacing x with -x.

First one could *almost* work, if there wasn't the minus sign at the beginning, so it's a no.

Second one is also a no, the original one is always positive, the second one is always negative.

3rd one is reflecting the graph all right, but it's also adding some sort of stretch (try with the point x=1 in both cases, $f(1) = \frac 32$ while $g_3(1) = 2/3$

4th one is the charm.

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N-4=16 solve equation

gizmo_the_mogwai [7]

n - 4 = 16 Given

+ 4 + 4 Cancel out 4 to isolate n variable

n = 20

Prove.

20 - 4 = 16

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True.

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3 years ago

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100 points Triangle ABC is shown below: Triangle ABC is shown. A is at 2, 1. B is at negative 1, 4. C is at 3, 7. If triangle AB

Vinil7 [7]

Using translation concepts, the coordinates for point C' are given as follows:

C'(-3,7).

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

The rule for a reflection over the y-axis is given as follows:

(x,y) -> (-x,y).

Hence, for point C', the coordinates are given by:

(3,7) -> (-3,7).

More can be learned about translation concepts at brainly.com/question/4521517

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2 years ago

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$