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3 years ago

PLEASE HELP ME I HAVE OTHER WORK TO DO PLEASE HELP omg

Mathematics

1 answer:

kolbaska11 [484]3 years ago

3 0

1. Decimal .25 percent 25%

Uhh I'm not good with this that's all I know‍♂️

You might be interested in

HELP!!

Leokris [45]

Answer:

1/2

Step-by-step explanation:

Given:

sinx=-1/2

cosy=√3/2

finding x

x=sin^-1(-1/2)

x=-π/6

x=-30°

finding y

cosy=√3/2

y=cos^-1(√3/2)

y=π/6

y=30°

Now finding cos(x-y)

cos(-π/6-π/6)

=cos(-π/3)

=1/2!

6 0

3 years ago

In order for the parallelogram to be a square, x = [?]. 4x + 17 12x - 23

AfilCa [17]

Answer:

x = 5

Step-by-step explanation:

4x + 17 = 12x - 23

23 + 17 = 12x - 4x

40 = 8x

8 8

5 = x

5 0

2 years ago

2. (-4) - 3- 1- 8 A) -10 B)-16 C) -19 D) -8

DedPeter [7]

Answer:

B) -16

Step-by-step explanation:

-4-3-1-8

subtract them all lol

7+9= 16

Then add the negative symbol

-16

7 0

3 years ago

Read 2 more answers

PLS HELP (the second page has the options)

Tom [10]

The 3rd answer because -3 is the smallest y intercept

6 0

3 years ago

Read 2 more answers

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

3 years ago