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2 years ago

What are the domain and range of the function below?

Mathematics

2 answers:

nignag [31]2 years ago

3 0

The domain and range is all real numbers.

<h3>Answer: D) Domain: (-∞, ∞); Range: (-∞, ∞)</h3>

Yakvenalex [24]2 years ago

3 0

Answer:

D) Domain: (-∞, ∞)

Range: (-∞, ∞)

Step-by-step explanation:

The domain is the input values, or the x values.

We can put in any x values for this function.

Domain : (-∞, ∞)

The range is the output values or the y values.

We can get any output values for this function

Range: (-∞, ∞)

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2 years ago

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3 years ago

Read 2 more answers

A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

TEA [102]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = $1,150

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

Let $\mu$ = mean of all account balances

(a)So, Null Hypothesis, $H_0$ : $\mu$ = $1,150 {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150 {means that the mean of all account balances is significantly different from $1,150}

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

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

where, $\bar X$ = sample average balance = $1,200

s = sample standard deviation = $126

n = sample of account balances = 81

(b) So, test statistics = $\frac{1,200-1,150}{\frac{126}{\sqrt{81} } }$ ~ $t_8_0$

= 3.571

The value of the sample test statistics is 3.571.

(c) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_8_0$ > 3.571) = Less than 0.05%

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0

3 years ago