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3 years ago

50 POINTS PLEASE ANSWER (correctly) FAST

Mathematics

1 answer:

ivolga24 [154]3 years ago

4 0

The graph of line g has an x-intercept of (2,0) and a y-intercept of (0,4).

Which statement(s) comparing these two functions is(are) true?

F. The y-value when x=−2 for line f is greater than the y-value when x=2 for line g.
A. The slope of line f is steeper than the slope of line g.
D. The y-intercept of line f is greater than the y-intercept of line g.

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Is 3.6 km or 3,600 m is greater or eqaul

SVEN [57.7K]

3.6 km is equal to 3,600 m :)

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4 years ago

Read 2 more answers

I NEED CORRECT ANSWER

RUDIKE [14]

Answer:

Step-by-step explanation:

Activity 3

Q1) consistent, independent

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4 0

3 years ago

6x^4-17x^3-7x^2+13x=3 what are all the solutions

wlad13 [49]

6x^4-17x^3-7x^2+13x=3
6x^4-17x^3-7x^2+13x-3=3-3
6x^4-17x^3-7x^2+13x-3=0
y=6x^4-17x^3-7x^2+13x-3
y=0

The solutions are: x=-1, 1/2, 1/3 and 3

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4 years ago

Evaluate the function for the given value.

maks197457 [2]

Answer:

$g(t) = 2 \cdot 0.4t \ for \ t = -2\\g(-2) = 2 * 0.4*-2 = -1.6$

none of the above.

Are u sure question is correct?

7 0

4 years ago

In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u

horsena [70]

Answer:

$\large\boxed{\large\boxed{0.119}}$

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

1. Start with the first 11 inspected packages:

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

$C(m,n)=\dfrac{m!}{m!(m-n)!}$

$C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}$

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

$C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}$

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

$C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}$

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

$C(7,3)\times C(13,8)$

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

$\dfrac{C(7,3)\times C(13,8)}{C(20,11)}$

Subsituting:

$\dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}$

$=\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885$

2. The 12th package

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

3. Finally

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

$0.26818885\times 4/9=0.119$

6 0

4 years ago