All categories

3 years ago

Find the value of f(7). y = f(x)

Mathematics

1 answer:

kozerog [31]3 years ago

8 0

Step-by-step explanation:

For x = 7 or y = f(7)

Graph and solution are in the picture below

You might be interested in

Gail purchase 5 raffle tickets. The grand prize is a vacation to Aruba. A total of 1,500 raffle tickets were sold. There is only

Aleks04 [339]

If it is in fraction you could do 5/1500 and reduce it to 1/300 , to change 1/300 to whole number it would be 3 , to change 1/300 to decimal it would be .03

8 0

3 years ago

Select the correct answer.

makvit [3.9K]

Answer:

Step-by-step explanation:

y = 52° ( Opposite Angles )

y + 2x = 180° ( Interior Angles )

$y + 2x = {180}^{o} \\ {52}^{o} + 2x = {180}^{o} \\ 2x = {180}^{o} - {52}^{o} \\ 2x = {128}^{o} \\ x = \frac{ {128}^{o} }{2} \\ x = {64}^{o} \\$

3 0

3 years ago

What is the midpoint between 0.13 and 0.14

weqwewe [10]

0.13 + 0.14= 0.27 0.27 ÷ 2 = 0.135 Therefore the midpt. Is 0.135

3 0

3 years ago

What is the approximate surface area of the sphere 16yd?

bazaltina [42]

314.159 and 5 radius r

8 0

3 years ago

Read 2 more answers

The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75

KatRina [158]

Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean:

$\mu = 47$

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

$4\sigma = 75 - 19$

$4\sigma = 56$

$\sigma = \frac{56}{4} = 14$

a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

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

$Z = \frac{25 - 47}{14}$

$Z = -1.57$

$Z = -1.57$ has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

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

$Z = \frac{60 - 47}{14}$

$Z = 0.93$

$Z = 0.93$ has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

3 0

3 years ago