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3 years ago

Could anyone help me?

Mathematics

2 answers:

egoroff_w [7]3 years ago

7 0

Answer:

The answer is C; 6

tatyana61 [14]3 years ago

3 0

C 6 I think I hope this helps

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3. Solve for y: -5x + 2y = 14

Schach [20]

Answer:

y= -1

Explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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2 years ago

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Mark bikes 4/10 of a mile in 1/5 of an hour. Find marks average speed in miles per hour.

Triss [41]

He bikes 4/10 of a mile in 1/5 of an hour

That is 4/10 miles in 1/5 hours

Average speed = Distance/ time = 4/10 ÷ 1/5 = 4/10 × 5/1 = 2 miles/hour

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3 years ago

Jin’s soccer team has 4 coaches and 22 players, and his twin sister’s team has 2 coaches and 11 players. If the relationship bet

nikklg [1K]

Question : Jin’s soccer team has 4 coaches and 22 players, and his twin sister’s team has 2 coaches and 11 players. If the relationship between the number of players and coaches is proportional throughout the league, which ordered pair could represent the total number of coaches and players in the entire league?

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Answer :

B :

26 coaches and 143 players

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3 years ago

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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.

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

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

A system of inequalities is graphed below. The variable x represents the number of cans someone is buying, and y represents the

8090 [49]

We have that
We will analyze each case to verify the answer
see the attached figure

Point A) 7 cans and 140 bottles------------> (x,y)=(7,140)----> is not solution
Point B) 12 cans and 40 bottles------------> (x,y)=(12,40)----> is not solution
Point C) 8 cans and 120 bottles------------> (x,y)=(8,120)----> is a solution
Point D) 9 cans and 70 bottles------------> (x,y)=(9,70)-------> is a solution

the answer is
the solutions are
8 cans and 120 bottles
9 cans and 70 bottles

6 0

3 years ago