SOMEONE HELP ME PLEASE

HEEELLLPPP PLEEAAASEEE

Pavel [41]

Answer:

B. y = 2.5

Step-by-step explanation:

Diane's average speed= 25/y km/h.

Ed's speed= (25/y) - 6 km/h.

Ed walks 25-3 = 22 km.

Consider Ed's journey:

Speed = distance / time so:

(25/y) - 6 = 22 / (y + 3) ( 3 hours longer = y + 3)

(25 /y - 6)(y + 3) = 22

25 + 75/y - 6y - 18 = 22

6y - 75/y = 25 - 18 - 22

6y - 75/y + 15 = 0

6y^2 + 15y - 75 = 0

2y^2 + 5y - 25 = 0

(2y - 5 )(y + 5) = 0

y = 2.5, -5.

y = 2.5 ( as it must be positive.

4 0

3 years ago

The area of a rectangle is 0.8 square units. The length is 3.2 units and the width is x units.

Mademuasel [1]

The answer is x=0.25

5 0

3 years ago

Solve the Quadratic equation

GrogVix [38]

Answer:

kindly using which rule?

7 0

3 years ago

Problem Solving: The function LaTeX: f\left(x\right)=50x+6500\:f ( x ) = 50 x + 6500represents the amount of money in a bank acc

Delicious77 [7]

Answer:

25x + 15800

Step-by-step explanation:

Given

The amount of money in both banks over time represented by the functions;

f ( x ) = 50 x + 6500 and g( x ) = − 25 x + 9300

The total amount of money in the two accounts over time is expressed as;

h(x)= f(x) + g(x)

Substitute;

h(x) = 50 x + 6500 + (− 25 x + 9300)

h(x) = 50x - 25x +6500+9300

h(x) =25x + 15800

Hence the sum of the money in the account over time is 25x + 15800

5 0

3 years ago

Elwin Osbourne, CIO at GFS, Inc., is studying employee use of GFS e-mail for non-business communications. A random sample of 200

ladessa [460]

Answer:

The 90% confidence interval for the population proportion is

(0.10872, 0.19128)

Step-by-step explanation:

Explanation:-

Step(i):-

Given data A random sample of 200 e-mail messages was selected. Thirty of the messages were not business related

Given random sample size 'n' = 200

Given Thirty of the messages were not business related

let 'x' = 30

Probability of the messages were not business related or proportion

$p = \frac{x}{n} = \frac{30}{200} = 0.15$

Step(ii):-

The 90% confidence interval for the population proportion is

$(p - Z_{0.10} \sqrt{\frac{p(1-p)}{n} } , p + Z_{0.10} \sqrt{\frac{p(1-p)}{n} } )$

Level of significance ∝ = 0.90 or 0.10

The critical value Z₀.₁₀ = 1.645

The 90% confidence interval for the population proportion is

$( 0.15-1.645 \sqrt{\frac{0.15(1-0.15)}{200} } , 0.15 +1.645 \sqrt{\frac{0.15(1-0.15)}{200} } )$

on calculation, we get

(0.15 - 0.04128 , (0.15 + 0.04128)

(0.10872, 0.19128)

Conclusion:-

The 90% confidence interval for the population proportion is

(0.10872, 0.19128)

5 0

3 years ago