Are the following triangles congruent? A) SAS B) ASA C) SSS D) NOT

zlopas [31]

If you would like to solve 2x + 5y = -13 and 3x - 4y = -8, you can do this using the following steps:

2x + 5y = -13 /*4
3x - 4y = -8 /*5
_______________
8x + 20y = -52
15x - 20y = -40
_______________
8x + 15x + 20y - 20y = -52 - 40
23x = -92
x = -92 / 23
x = -4

3x - 4y = -8
3 * (-4) - 4y = -8
-12 - 4y = -8
12 + 4y = 8
4y = 8 -12
4y = -4
y = -1

The correct result would be x = -4 and y = -1.

6 0

2 years ago

Please answer using the information provided (30 points for brainlyest)

Aleks [24]

$8.5(3.5 + x)$

$(8.5 \times 3.5) + (8.5 \times x)$

$29.75 + 8.5x$

I honestly don't know what to do for b. Maybe it tells us that the minimum area possible is 38.25 in²? or maybe it tells us that the area of just the original photo is 29.75 in²? I don't know.

7 0

2 years ago

As a manager for an advertising company, you must plan a campaign designed to increase Twitter usage. A recent survey suggests t

kolbaska11 [484]

Given:
Sample proportion = 85% = 0.85.
Confidence level (CL) = 90%.
Confidence interval (CI) $\hat{p} \pm z^{*} \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \\ where \\ n=sample\, size \\ z^{*}=1.645 \, at \, 90\% \, CL$

We want the confidence interval to be 5% or 0.05. Therefore
$0.85 - 1.645 \sqrt{ \frac{0.85(1-0.85)}{n} }=0.05 \\ 0.85 - \frac{0.5874}{ \sqrt{n} } =0.05 \\ \frac{0.5874}{ \sqrt{n} }=0.8 \\ \sqrt{n} = .7342 \\ n=0.54$

Because this number is less than 1, a default sample size of 30 (for the normal distribution) is recommended.

Answer: 30

3 0

3 years ago

You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte

Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

$f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.$

a) We calculate the probability that you need less than 5 minutes to get up:

$P(x$

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate the probability that you will need between 8 and 13 minutes:

$P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3$

Therefore, the probability is P(8<x<13)=0.3.

d) We calculate the probability that you will be late to each of the 9:30am classes next week:

$P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6$

You have 9:30am classes three times a week. So, we get:

$P=0.6^3=0.216$

Therefore, the probability is P=0.216.

e) We calculate the probability that you are late to at least one 9am class next week:

$P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1$

Therefore, the probability is P=1.

3 0

2 years ago

Jack and Jill are brother and sister. Jack rakes the yard in 5 hours and Jill rakes the yeard in 8 hours. Mom says to rake the y

SVETLANKA909090 [29]

Given that Jack rakes the yard in 5 hours and Jill rakes the yeard in 8 hours. Let the amount of time that it takes them to rake the yard together be t, then:

$\frac{1}{t} = \frac{1}{5} + \frac{1}{8} \\ \\ = \frac{8+5}{40} = \frac{13}{40} \\ \\ \therefore t= \frac{40}{13} =3 \frac{1}{13}\ hours$

8 0

3 years ago