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3 years ago

which property is illustrated by the equation below (r+s)+t=r+(s+t). A)Associative B)Commutative C)Distributive D)Identity

1 answer:

4 0

The associative property of addition is being demonstrated here. The parenthesis are moved around to "associate" or "group" the values into two different pairs.

Note: the associative property of multiplication is very similar and it is
a*(b*c) = (a*b)*c

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I'm doing Quadratic formula and I've simplified it down to the equation at the bottom. What do I do after? Is that my answer bec

givi [52]

No you could have left it as the last but one line or write them separately as

11/18 + √85 / 18 , 11/18 - √85/18

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3 years ago

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shutvik [7]

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3 years ago

Find the equation of the graphed line.

GrogVix [38]

Answer:

a. y = -1/2x - 2

Step-by-step explanation:

The correct answer choice can be determined by finding the slope of the line. The slope is the ratio of rise to run.

<h3>slope</h3>

The x-intercept is 4 units left of the y-axis. As the line "runs" those 4 units, it "rises" -2 units to intercept the y-axis at -2. The slope of the line is ...

m = rise/run = -2/4 = -1/2

In the slope-intercept form of the equation of a line, the slope is the coefficient of x. This information is sufficient to let us choose the first answer choice.

<h3>equation</h3>

The slope-intercept equation is ...

y = mx +b . . . . . . . slope m, y-intercept b

We know the slope is -1/2, and the y-intercept is where x=0, at y=-2. Then the equation is ...

y = -1/2x -2 . . . . . matches choice A

_____

Additional comment

When answering multiple-choice questions, you only need to do enough work to tell which answers are not viable.

When we plot the points, we see that the line has negative slope. (eliminates choice C). The slope is shallow, rather than steep (the x-intercept is farther from the origin than the y-intercept), so the magnitude of the slope is less than 1 and choices B and D are eliminated.

8 0

2 years ago

Perform the indicated operation: (x 2 -5x+2)+( x 2 +4x+5)-( x 2 -6x-42)

nordsb [41]

I think its (x^2)+5x+49

6 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago