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REY [17]
3 years ago
9

Graph each absolute value function. State the domain, range, and y-intercept.

Mathematics
1 answer:
lions [1.4K]3 years ago
5 0

Answer:

I. D: x\in R

ii. R: y\le0

iii. Y-int: b=-6

Step-by-step explanation:

i) The given absolute value function is

y=-3|x+2|

The domain is all values of x that will make the function defined.

The absolute value function is defined for all real numbers.

The domain is all real numbers.

ii) The range is the values of y for which x is defined.

The given absolute value function is

y=-3|x+2|

This function has the vertex at (-2,0) and it is reflected in the x-axis therefore the vertex is the maximum point on the graph.

The y-value of the vertex is the maximum value on the graph.

Hence the range is y\le0

iii) To find the y-intercept, we put x=0, into the function to get;

y=-3|0+2|

y=-3|2|

y=-3(2)

y=-6

The y-intercept is b=-6 or (0,-6)

See graph in attachment.

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7 0
3 years ago
Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall th
Anit [1.1K]

Answer:

The 99% confidence interval is (3.0493, 3.4907).

We are 99% sure that the true mean of the students Perry score is in the above interval.

Step-by-step explanation:

Our sample size is 21.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 21-1 = 20.

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 20 and 0.005 in the two-sided t-distribution table, we have T = 2.528

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

s = \frac{0.40}{\sqrt{21}} = 0.0873

Now, we multiply T and s

M = 2.528*0.0873 = 0.2207

Then

The lower end of the interval is the mean subtracted by M. So:

L = 3.27 - 0.2207 = 3.0493

The upper end of the interval is the mean added to M. So:

LCL = 3.27 + 0.2207 = 3.4907

The 99% confidence interval is (3.0493, 3.4907).

Interpretation:

We are 99% sure that the true mean of the students Perry score is in the above interval.

7 0
3 years ago
Select the correct answer.
Nina [5.8K]

Answer:

D. x^2 + 3x - 88

Step-by-step explanation:

Area of rectangle = length (l) * width (w)

Length of rectangle = (x - 8)

Width = (x + 11)

Area of rectangle = (x - 8)(x + 11)

Expand the expression

x(x + 11) -8(x + 11) (distributive property of multiplication)

x^2 + 11x - 8x - 88

Combine like terms

x^2 + 3x - 88

Expression for the area of the rectangle = x^2 + 3x - 88

7 0
3 years ago
For EACH ordered pair, determine if its a solution to 2x+9y=13
vladimir1956 [14]
The answer is (2,1). 2*2=4 and 9*1=9. 9+4=13
8 0
3 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0
3 years ago
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