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nikitadnepr [17]

3 years ago

6

Classify the following triangle. Check all that apply

2 answers:

pogonyaev3 years ago

8 0

Its B and D thats all i know about it you can trust this answer.

kramer3 years ago

7 0

D and E. If I remember right that should be the answer.

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Please help! I dont understand! Thank you

Digiron [165]

QUESTION A

The given inequality is

$x-1\:>\:5$

Add 1 to both sides of the inequality to obtain;

$x\:>\:5+1$

$x\:>\:6$

Answer: 2. $x\:>\:6$ , open dot at 6 and shaded to the right.(A)

QUESTION B.

The given inequality is $2x \le 6$ .

Divide both sides by 2;

$x \le \frac{6}{2}$ .

$x \le 3$ .

Answer: 5. $x \le 3$ , closed dot at 3 and shaded to the left(B).

QUESTION C

The given inequality is $x-7\:>\:-4$

Add 7 to both sides to get;

$x\:>\:-4+7$

$\Rightarrow x\:>\:3$

Answer: 7. $x\:>\:3$ , open dot at 3 and shaded to the right.(C)

QUESTION D

We want to solve;

$-2x \:$

Divide through by -2 and reverse the inequality sign;

$x \:>\:\frac{6}{-2}$

$x \:>\:-3$

Answer; 3. $x \:>\:-3$ Open dot at -3 and shaded to the right. (D)

QUESTION E

We have

$4\:$

Divide through by -4 and reverse the sign.

$\frac{4}{-4}\:>\:x$

$-1\:>\:x$

or

$x\:$

Answer: 6. $x\:$ open dot at -1, and shaded to the left(E)

QUESTION F

The given inequality is $-2x+3\:$

Add -3 to both sides to get;

$-2x\:$

$-2x\:$

Divide both sides of the inequality by -2 and reverse the sign;

$x\:>\:5$

Answer:9. $x\:>\:5$ , open dot at 5 and shaded to the right(F)

QUESTION G

The given inequality is

$5x+1\ge11$

$5x\ge11-1$

$5x\ge10$

Divide by 5

$x\ge2$

Answer ; 10. $x\ge2$ , closed dot at 2 and shaded to the right(G)

QUESTION H

We have $3(x+4)\:>\:8x-6$

Expand the bracket to get;

$3x+12\:>\:8x-6$

Group similar terms to get;

$3x-8x\:>\:-6-12$

Simplify

$-5x\:>\:-18$

Divide through by -5 and reverse the sign;

$x\:$

Answer: 1. $x\:$ Open dot at 18/5 and shaded to the left(H)

QUESTION I

We have $-3x+4\:$ .

$-3x+x\:$

$-2x\:$

Divide by -2 and reverse the sign;

$x\:>\:1$

Answer: 8. $x\:>\:1$ , open dot at 1 and shaded to the right(I)

J. The given inequality $-2(x-4)\:>\:5-(x+2)$

Expand the bracket to get;

$-2x+8\:>\:5-x-2$

Group like terms to get;

$-2x+x\:>\:5-2-8$

$-x\:>\:-5$

Divide by -1 and reverse the sign to get;

$x\:$

Answer; 4. $x\:$ , open dot at 5 and shaded to the left.(J)

6 0

3 years ago

To increase an amount by 7%, what single multiplier would you use ?

Scrat [10]

You would use 1.07 for that

6 0

3 years ago

Read 2 more answers

It takes a computer 1.2 seconds to perform an operation. If a task contains eight operations how many seconds will it take for c

serg [7]

Answer:

19.2 seconds

Step-by-step explanation:

1.2 times 8 which equals 9.6

9.6 times 2

7 0

3 years ago

Read 2 more answers

Round 62.85 to the nearest tenth

Fiesta28 [93]

Answer:

62.9

Step-by-step explanation:

Follow the rounding rules.

Remember the tenths place is one place after the decimal point.

5 0

3 years ago

Read 2 more answers

A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores

EastWind [94]

Answer:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

Step-by-step explanation:

We have the following info given from the problem

$\bar X_1 = 52$ sample mean for this year

$s_1= 13$ sample deviation for this year

$n_1 = 75$ random sample selected for this year

$\bar X_2 = 49$ sample mean for last year

$s_2= 11$ sample deviation for last year

$n_1 = 53$ random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given by:

$df= n_1 +n_2 -2 = 75+53-2= 126$

The confidence level is 0.95 and the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ so then the critical value for this case is :

$t_{\alpha/2}= 1.979$

The margin of error would be:

$ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300$

And the confidence interval would be given by:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

6 0

3 years ago