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3 years ago

Given: g ∥ h and ∠2 ≅ ∠3

Mathematics

2 answers:

Yuliya22 [10]3 years ago

7 0

Answer: What's up?

LOL

nordsb [41]3 years ago

3 0

Answer:3

Step-by-step explanation:

im pretty sure give brainliest

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6x+5y=-15 into intercept form

Yuki888 [10]

$General\ equation\ for\ line\ in\ slope\ intercept\ form:\\\\y=ax+b\\\\6x+5y=-15\\
5y=-15-6x \ \ |:5\\
y=-3-\frac{6}{5}x\\\\
In\ slope\ intercept\ form\ y=-3-\frac{6}{5}x.$

5 0

3 years ago

Read 2 more answers

A line intersects the points (13,-4) and (1,12).Find the slope and simplify completely

Dvinal [7]

Answer:

4/-3

Step-by-step explanation:

To find the slope

y2 - y1 / x2 - x1 =

12 - -4 16

---------- = ----

1 - 13 -12

Simplify by dividing both parts by 4

4/-3 is your slope :)

6 0

3 years ago

Plutonium–238 has a yearly decay constant of 7.9 × 10-3. If an original sample has a mass of 15 grams, how long will it take to

eduard

The answer is 28 years

At = A0 * e^(-k * t)
At = 12 g
A0 = 15 g
k = 7.9 × 10^-3 = 0.0079
t = ?

12 = 15 * e^(-0.0079 * t)
12/15 = e^(-0.0079 * t)
0.8 = e^(-0.0079 * t)

Logarithm both sides (because ln(e) = 1:
ln(0.8) = ln(e^(-0.0079 * t))
ln(0.8) = (-0.0079 * t) * ln(e)
-0.223 = -0.0079 * t
t = -0.223 / -0.0079
t = 28.23
t ≈ 28 years

8 0

3 years ago

What is 574 ÷ 82 my math say use compatible numbers to round first then awnser the actual queston

Svetradugi [14.3K]

Answer: 7 explanation:574 $%$ %82=7 7x82=574

7 0

3 years ago

Read 2 more answers

At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$

$P( X \le 4 ) = 0.0054$

4 0

3 years ago