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2 years ago

14

PLEASE HELP WITH QUICK QUESTION! 25 POINTS!

2 answers:

Damm [24]2 years ago

6 0

Isn’t it a 90 degree angle

Bas_tet [7]2 years ago

3 0

Answer:

90 degree angle

Step-by-step explanation:

You might be interested in

Solve.<br><br> x^-9 y^3 / x^-7 y^8

aniked [119]

$\dfrac{x^{-9}y^3}{x^{-7}y^8}=x^{-9-(-7)}y^{3-8}=x^{-2}y^{-5}=\dfrac{1}{x^2y^5}\\\\Used:\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\a^{-n}=\dfrac{1}{a^n}$

3 0

3 years ago

What is the mean?<br> 1 6 1 10 5 5 7

algol13

Answer:

5

Step-by-step explanation:

if u add al of these it equals 35 divided by 7 equals 5

7 0

2 years ago

4 people went to the movies. Each person spent $9.50 on a ticket and had the same amount of money leftover. They originally had

solong [7]

$2.50 was left for each person.

4(9.50) + x = 48

38 + x = 48

subtract 38 on both sides

x = 10

10$ was left in total by all 4 people

10/4 = 2.50

$2.50 was left over for each person

7 0

3 years ago

Explain whether the following given sets are closed under addition:

Ivan

Answer:

Step-by-step explanation:

a) Yes.

$\forall\ a,b\ \in\ B,\ a+b\in B\ (since\ 0+0=0)$

b) Yes

$W=\{0,1,2,3,..,\}\ (whole\ numbers)\\\forall\ a,b\ \in\ 3W,\ a+b\in 3W\\since:\\a\ \in\ 3W \Longrightarrow\ \exists\ k_1\in W | a=3*k_1\\\\b\ \in\ 3W \Longrightarrow\ \exists\ k_2\in W | b=3*k_2\\\\a+b=3*k_1+3*k_2=3*(k_1+k_2) \in 3W\\$

c) Yes

$\forall n\ \in\ \mathbb{N}: \Longrightarrow\ n+1 \in\ \mathbb{N}\\$

d) No

$3\in V , 3+3=6\notin V\\$

7 0

2 years ago

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

natulia [17]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

2 years ago