What is the equation of the line in slope-intercept form that passes through (-5, 3) and (2, -4)?

Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mea

lord [1]

Answer:

a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

$\int\limits^y_0 {kx^{2} } \, =1$

where y=23

Clearing k=0.00025

a) $Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5$

b) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2} =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6$

c) The function is equal to:

f(x)=k(1+2x)

$\int\limits^y_0 {k(1+2x)} \, =1$

where y=20

k=0.0024

$Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3$

d) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2} =198.4-176.89=21.51$

8 0

3 years ago

A clock was reading the time accurately on Friday at noon. On Monday at 6pm the clock was running late by 468 seconds. On averag

Setler [38]

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × $\frac{1}{2}$ = 24

Saturday = 48

Sunday = 48

Monday = 48 × $\frac{3}{4}$ = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

To learn more about clock problems visit:

brainly.com/question/27122093.

#SPJ1

3 0

2 years ago

May someone please help me with 1-5? I have been stuck on this for a while now.

AleksAgata [21]

Answer:

1 and 2 are correct!! Good Job!!

3. on the number line put a solid dot over 0, the line should to the right.

4. on the number line put a solid dot over -4, the line will go to the left.

5. on the number line put a open dot over 1.5, the line will go to the left.

Step-by-step explanation:

I hope this helps!! Also, you did great on 1 and 2, got this!!

5 0

3 years ago

You must include an explanation accompanied with your solution. Answers without

nekit [7.7K]

Answer:

B) $7.50

Step-by-step explanation:

Interest (I) = Principal (P) x Interest Rate (r) x Time (t, in years)

I = 125(.03)(2)

I = 7.5

4 0

2 years ago

The perimeter of a rectangle is 15x + 17 y if the length is 7/2x + 7y then find the width of the rectangle A) 4x + 3/2y B) 8x +

tigry1 [53]

The equation for the perimeter of a rectangle:

P = 2L + 2W

We’re given P and L and asked to find W:

15x + 17y = 2(7/2x + 7y) + 2W
15x + 17y = 7x + 14y + 2W
2W = 8x + 3y
W = 4x + 3/2y

Answer is A.

3 0

4 years ago