All categories

3 years ago

What is the domain and range?

Mathematics

1 answer:

MrRissso [65]3 years ago

7 0

Hello from MrBillDoesMath!

Answer:

Domain = { 5. 10. 15. 20}

Range = -3,-1,1,3} .

Discussion:

The domain is the set of all "x" values, that is { 5. 10. 15. 20} and the range is (-3,-1,1,3} . The arrows in the diagram show the mapping between the domain and the range. For examples, 5 -> 1 ( "x value 5 is mapped to y value 1"), 10 ->3, 15 ->-3, and 20 -> -2

Thank you,

MrB

You might be interested in

What is the slope of this table?

spayn [35]

The would be -3,, because your doing y2-y1 over x2-x1

7 0

3 years ago

A quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain bra

tensa zangetsu [6.8K]

Answer:

We conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

Step-by-step explanation:

We are given that a quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain brand of soft drink contains 120 calories as the labeling indicates.

Using a random sample of 10 cans, the manager determined that the average calories per can is 124 with a standard deviation of 6 calories.

Let $\mu$ = average calorie content of a 12-ounce can.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 120 calories {means that the average calorie content of a 12-ounce can is less than or equal to 120 calories}

Alternate Hypothesis, $H_A$ : $\mu$ > 120 calories {means that the average calorie content of a 12-ounce can is greater than 120 calories}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average calories per can = 124 calories

s = sample standard deviation = 6 calories

n = sample of cans = 10

So, test statistics = $\frac{124-120}{\frac{6}{\sqrt{10} } }$ ~ $t_9$

= 2.108

The value of t test statistics is 2.108.

Now, at 0.05 significance level the t table gives critical value of 1.833 at 9 degree of freedom for right-tailed test. Since our test statistics is more than the critical values of t as 2.108 > 1.833, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

7 0

3 years ago

All of the following are equivalent except___. 7x^3,4x+3x,(4+3)x,7x

Paha777 [63]

7x^3 is not equivalent to the other varieties of 7x

6 0

4 years ago

Read 2 more answers

In a class of 6, there are 2 students who forgot their lunch.

const2013 [10]

The probability is 1/2

5 0

3 years ago

Heights of men on a baseball team have a bell-shaped distribution with a mean of 181 cmand a standard deviation of 6 cm.Using t

Ede4ka [16]

Answer:

a) 95.4% of the men are between 169cm and 193cm

b) 68.2% of the men are between 175cm and 187cm

Step-by-step explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the Heights of men on a baseball team.

From the problem we have the mean and the standard deviation for the random variable X. $E(X)=181, Sd(X)=6$

So we can assume $\mu=181 , \sigma=6$

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68 2

• The probability of obtain values within two deviation's from the mean is 0.95 4

• The probability of obtain values within three deviation's from the mean is 0.997

a. 169cm and 193cm

We can find the deviations from the mean using the z score formula given by:

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

IF we find the z score for 169 and 193 we got:

$Z=\frac{169-181}{6}=-2$

$Z=\frac{193-181}{6}=2$

Since we are within 2 deviations from the mean and using the empirical rule we have 95.4% of the data

95.4% of the men are between 169cm and 193cm

b. 175cm and 187cm

We can find the deviations from the mean using the z score formula given by:

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

IF we find the z score for 169 and 193 we got:

$Z=\frac{175-181}{6}=-1$

$Z=\frac{187-181}{6}=1$

Since we are within 1 deviations from the mean and using the empirical rule we have 68.2% of the data

68.2% of the men are between 175cm and 187cm

5 0

3 years ago