Please help with the statistics

Please solve using the image. No links.

Nesterboy [21]

Answer:

1/10 + 1/2 = 1/10 + 5/10 = 6/10

6 0

3 years ago

Read 2 more answers

Fine the equation of the line through (o, -3) perpendicular to y =4x+4

svlad2 [7]

Answer:

The equation of the line is $y = -\frac{x}{4} - 3$

Step-by-step explanation:

Equation of a line:

The equation of a line has the following format:

$y = mx + b$

In which m is the slope and b is the y-intercept(the value of y when x = 0).

Perpendicular lines:

If two lines are perpendicular, the multiplication of their slopes is -1.

Line perpendicular to y = 4x + 4.

This line has slope 4, so our line will have slope:

$4m = -1$

$m = -\frac{1}{4}$

So

$y = -\frac{x}{4} + b$

Through (0, -3)

This means that $b = -3$

So

$y = -\frac{x}{4} - 3$

8 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago

A deposited 7500 Dollars in a bank and received interest of 900 Dollars after one year. B received interest of 1440 Dollars afte

White raven [17]

Answer:

If the rate of interest is 12% than the answer is 12000

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

7 0

2 years ago

Jesse is investing $2000 more in stocks than in bonds.The bonds pay 7.2% in interest and the stocks pay 6%. If the income from e

tamaranim1 [39]

X+2,000=stocks
x=bonds
7.2%⇒(.072)(x)=bonds invested
6%⇒(.06)(x+2,000)=stocks invested

Here is the equation:

x(.072)=(.06)(x+2,000) | Since the income from both equal

You then solve each side first (assume you know how to but if not, let me know)

You first multily the parentheses/x variables
⇒.072x = .06x+2,000(.06)

⇒.072x = .06x+120

You then minus .06x from each side
⇒.012x = 120

You then divide each side by .012
.012x÷.012=x
120÷.012=10,000
x=10,000

Now change the original x variables with 10,000

.072($10,000)=$720 bond interest
They are equal so $720×2=$1440

7 0

3 years ago