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2 years ago

Find a linear inequality with the following solution set. Each grid line represents one unit.

Mathematics

1 answer:

ikadub [295]2 years ago

5 0

The linear inequality in standard form is: x + y < -2 or x + y + 2 < 0

<h3>How to Write a Linear Inequality?</h3>

To determine what sign to use, follow these rules:

Use "<" or ">" when the boundary line is dotted or dashed.
Use "<" or "≤" when the shaded area is under the boundary line (dotted or dashed).
Use "≤" or "≥" when the boundary line is not dotted or dashed.
Use ">" or "≥" when the shaded area is above the boundary line (dotted or dashed).

The graph given has the following features:

It has a boundary line that is dotted/dashed.
It has a shaded area that is under the boundary line.
Thus, the inequality sign we would use is, "<".

Find the slope:

Slope (m)= rise/run = -(2 units) / (2 units).

Slope (m) = -1.

The y-intercept (b) = -2.

Substitute m = -1 and b = -2 into y < mx + b:

y < (-1)x + (-2)

y < -x - 2

Rewrite

x + y < -2

x + y + 2 < 0

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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3 years ago

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4 years ago

Classify the following triangle as acute, obtuse, or right.<br>O A. Acute<br>OB. Obtuse<br>C. Right

makvit [3.9K]

Answer:

right

Step-by-step explanation: