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3 years ago

Which table shows the reflection of f(x) across the y-axis? HELP ASAP!!

Mathematics

1 answer:

tangare [24]3 years ago

5 0

Answer:

The first one

Step-by-step explanation:

When you do a reflection across the y-axis, it is like a mirror. And if you have a generic point (x, y) the new point will be (-x, y).

In the table you have 4 points:

(x,y)

(-3,9)

(0,0)

(2,4)

(5,25)

The new points will be:

(3,9)

(0,0)

(-2,4)

(-5,25)

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The number of people using an older version of a spreadsheet program decreases at a rate that is proportional at any time to the

Paraphin [41]

Answer:

5842 people after 2 months

Step-by-step explanation:

Given

Initial Value: $a =10000$

Rate: $r = 20\%$ every 6 months

Required

Number of people using it after 2 months

The function follows an exponential model.

So, we need to first write out the function.

If Rate: $r = 20\%$ every 6 months

Then

Rate: $r = 40\%$ after 1 year

$r = 0.04$

The function is represented as:

$y = ar^x$

So:

$y = 10000 * 0.04^t$

For 2 months, we have:

$t = 2\ months$

Convert to years

$t = \frac{2}{12}$

$t = 0.167$

$y = 10000 * 0.04^t$

$y = 10000 * 0.04^{0.167}$

$y = 5842$

8 0

3 years ago

I DID NOT MEAN TO POST

Nataliya [291]

Okie thank you you you

6 0

3 years ago

What is the quotient of 5 1/2 and 1/8

Stels [109]

Answer:

Step-by-step explanation:

5 1/2 / 1/8

turn into improper fraction

11/2 then change to multiplication

11/2*8/1(have to flip)

88/2(multiply across)

simplify=44

4 0

3 years ago

3x +12 =90 what is x

castortr0y [4]

Answer:

x=26

Step-by-step explanation:

3x +12 =90

subtract 12 from each side

3x=78

divide 3 from each side

x=26

3 0

3 years ago

Read 2 more answers

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago