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Alexxandr [17]
3 years ago
10

Use the graph of y =g(x) to graph the given function.

Mathematics
1 answer:
Semenov [28]3 years ago
3 0

Answer:

See image attached

Step-by-step explanation:

This type of transformation corresponds to a reflection of the function's graph around the x axis. Therefore one should obtain the mirror image of the segments on he upper quadrant as shown in red in the attached image.

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P/-8 + 9 &gt; 13 <br> pls explain i don't get it
KonstantinChe [14]

Answer:

p < -32

Step-by-step explanation:

<em>Given:</em> p/-8 + 9 > 13

<em>Rewrite:</em> \frac{p}{-8} + 9 > 13

<em>Subtract 9 from both sides:</em> \frac{p}{-8}  > 4

<em>Move the negative (it is a fraction): </em>\frac{-p}{8} > 4

<em>Multiply both sides by 8:</em> -p > 32

<em>Times by -1 (-p -> -1p): </em>p > -32

<em>Flip the sign (multiplied by a negative): </em>p < -32

Hope this helps, have a nice day :D

3 0
2 years ago
Read 2 more answers
Color
Anit [1.1K]

Answer:

Yellow.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes. Experimental probability is found taking into account previous experiments.

In this question:

6 + 5 + 4 + 10 = 25 total marbles.

The experimental probabilities are:

6/25 probability it is red.

5/25 = 1/5 probability it is yellow. Thus, the answer is yellow.

4/25 probability it is blue.

10/25 = 2/5 probability it is green.

5 0
2 years ago
Can yall help answer this asap​
Trava [24]

Answer: B

(press thanks please) i need it lol. . . . . . . . . . . . . . .

8 0
3 years ago
Ms. Moore has a garden in the shape of a rectangle with width measuring 8 km and a length measuring 14 km. What is the area of t
iVinArrow [24]
Answer: A = 112km^2

Explanation:

A = L x W
A = 14km x 8km
A = 112km^2
7 0
3 years ago
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.6 ppm and standard de
Setler79 [48]

We assume that question b is asking for the distribution of \\ \overline{x}, that is, the distribution for the average amount of pollutants.

Answer:

a. The distribution of X is a normal distribution \\ X \sim N(8.6, 1.3).

b. The distribution for the average amount of pollutants is \\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}}).

c. \\ P(z>-0.08) = 0.5319.

d. \\ P(z>-0.47) = 0.6808.

e. We do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for \\ \overline{X} is also normal because <em>the sample was taken from a normal distribution</em>.

f. \\ IQR = 0.2868 ppm. \\ Q1 = 8.4566 ppm and \\ Q3 = 8.7434 ppm.

Step-by-step explanation:

First, we have all this information from the question:

  • The random variable here, X, is the number of pollutants that are found in waterways near large cities.
  • This variable is <em>normally distributed</em>, with parameters:
  • \\ \mu = 8.6 ppm.
  • \\ \sigma = 1.3 ppm.
  • There is a sample of size, \\ n = 38 taken from this normal distribution.

a. What is the distribution of X?

The distribution of X is the normal (or Gaussian) distribution. X (uppercase) is the random variable, and follows a normal distribution with \\ \mu = 8.6 ppm and \\ \sigma =1.3 ppm or \\ X \sim N(8.6, 1.3).

b. What is the distribution of \\ \overline{x}?

The distribution for \\ \overline{x} is \\ N(\mu, \frac{\sigma}{\sqrt{n}}), i.e., the distribution for the sampling distribution of the means follows a normal distribution:

\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}}).

c. What is the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants?

Notice that the question is asking for the random variable X (and not \\ \overline{x}). Then, we can use a <em>standardized value</em> or <em>z-score</em> so that we can consult the <em>standard normal table</em>.

\\ z = \frac{x - \mu}{\sigma} [1]

x = 8.5 ppm and the question is about \\ P(x>8.5)=?  

Using [1]

\\ z = \frac{8.5 - 8.6}{1.3}

\\ z = \frac{-0.1}{1.3}

\\ z = -0.07692 \approx -0.08 (standard normal table has entries for two decimals places for z).

For \\ z = -0.08, is \\ P(z.

But, we are asked for \\ P(z>-0.08) \approx P(x>8.5).

\\ P(z-0.08) = 1

\\ P(z>-0.08) = 1 - P(z

\\ P(z>-0.08) = 0.5319

Thus, "the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants" is \\ P(z>-0.08) = 0.5319.

d. For the 38 cities, find the probability that the average amount of pollutants is more than 8.5 ppm.

Or \\ P(\overline{x} > 8.5)ppm?

This random variable follows a standardized random variable normally distributed, i.e. \\ Z \sim N(0, 1):

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

\\ z = \frac{\overline{8.5} - 8.6}{\frac{1.3}{\sqrt{38}}}

\\ z = \frac{-0.1}{0.21088}

\\ z = \frac{-0.1}{0.21088} \approx -0.47420 \approx -0.47

\\ P(z

Again, we are asked for \\ P(z>-0.47), then

\\ P(z>-0.47) = 1 - P(z

\\ P(z>-0.47) = 1 - 0.3192

\\ P(z>-0.47) = 0.6808

Then, the probability that the average amount of pollutants is more than 8.5 ppm for the 38 cities is \\ P(z>-0.47) = 0.6808.

e. For part d), is the assumption that the distribution is normal necessary?

For this question, we do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for \\ \overline{X} is also normal because the sample was taken from a normal distribution. Additionally, the sample size is large enough to show a bell-shaped distribution.  

f. Find the IQR for the average of 38 cities.

We must find the first quartile (25th percentile), and the third quartile (75th percentile). For \\ P(z, \\ z \approx -0.68, then, using [2]:

\\ -0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}

\\ (-0.68 *0.21088) + 8.6 = \overline{X}

\\ \overline{x} =8.4566

\\ Q1 = 8.4566 ppm.

For Q3

\\ 0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}

\\ (0.68 *0.21088) + 8.6 = \overline{X}

\\ \overline{x} =8.7434

\\ Q3 = 8.7434 ppm.

\\ IQR = Q3-Q1 = 8.7434 - 8.4566 = 0.2868 ppm

Therefore, the IQR for the average of 38 cities is \\ IQR = 0.2868 ppm. \\ Q1 = 8.4566 ppm and \\ Q3 = 8.7434 ppm.

4 0
3 years ago
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