Y=-2/5x a. direct b. inverse c. neither

Simplify the expression: –7c8·(–0.4c3)2

MArishka [77]

Answer:

A simplified version of “–7c8·(–0.4c3)2,” would be,

“–7c8·(–0.4c3)²= (-7c8)(-0.4)²(c)^6”

If it helped please mark as brainliest, thanks. :)

4 0

3 years ago

jessa jane, and dina bought materials for thir project in math worth 306.75 pesos the girls divided the amount equally among the

Yuki888 [10]

Answer:

yeet

Step-by-step explanation:

ya dig

8 0

2 years ago

The maker of an automobile advertises that it takes 12 seconds to accelerate from 20 kilometers per hour to 65 kilometers per ho

sp2606 [1]

Answer:

The acceleration is 1.0416 m/ $s^{2}$

Step-by-step explanation:

In order to solve this problem we first need to know the formula for acceleration which is the following.

$acceleration = \frac{final.velocity - initial.velocity}{final.time - initial.time}$

Since the time acceleration is calculated as $m/s^{2}$ we need to convert the km/h into m/s. Since 1km = 1000m and 1 hour = 3600 seconds, then

$\frac{20*1000 }{3600s} = \frac{20,000m}{3600s} = \frac{20m}{3.6s}$

**Dividing numerator and denominator by 1000 to simplify**

$\frac{65*1000 }{3600s} = \frac{65,000m}{3600s} = \frac{65m}{3.6s}$

**Dividing numerator and denominator by 1000 to simplify**

Now we can plug in the values into the acceleration formula to calculate the acceleration.

$acceleration = \frac{\frac{65m}{3.6s}-\frac{20m}{3.6s} }{12s-0s}$

$acceleration = \frac{\frac{45m}{3.6s}}{12s}$

$acceleration = \frac{\frac{12.5m}{s}}{12s}$

$acceleration = \frac{\frac{1.0416m}{s}}{s}$

Finally we can see that the acceleration is 1.0416 m/ $s^{2}$

8 0

3 years ago

Determine whether the equation below has a one solutions, no solutions, or an infinite number of solutions. Afterwards, determin

egoroff_w [7]

Answer:

no solution

Step-by-step explanation:

X cannot equal itself minus 5, in the same way if we replace it by let's say, 6...

6 does not equal 6-5 (1)........ and by the way in this specific equation 6 is just an example it does not matter what you replace it with, x can never = x-5

Hope this helped! Good Luck!

5 0

4 years ago

A broker has calculated the expected values of two different financial instruments X and Y. Suppose that E(x)= $100, E(y)=$90 SD

Sveta_85 [38]

Expectation is linear, meaning

E(a X + b Y) = E(a X) + E(b Y)

= a E(X) + b E(Y)

If X = 1 and Y = 0, we see that the expectation of a constant, E(a), is equal to the constant, a.

Use this property to compute the expectations:

E(X + 10) = E(X) + E(10) = $110

E(5Y) = 5 E(Y) = $450

E(X + Y) = E(X) + E(Y) = $190

Variance has a similar property:

V(a X + b Y) = V(a X) + V(b Y) + Cov(X, Y)

= a^2 V(X) + b^2 V(Y) + Cov(X, Y)

where "Cov" denotes covariance, defined by

E[(X - E(X))(Y - E(Y))] = E(X Y) - E(X) E(Y)

Without knowing the expectation of X Y, we can't determine the covariance and thus variance of the expression a X + b Y.

However, if X and Y are independent, then E(X Y) = E(X) E(Y), which makes the covariance vanish, so that

V(a X + b Y) = a^2 V(X) + b^2 V(Y)

and this is the assumption we have to make to find the standard deviations (which is the square root of the variance).

Also, variance is defined as

V(X) = E[(X - E(X))^2] = E(X^2) - E(X)^2

and it follows from this that, if X is a constant, say a, then

V(a) = E(a^2) - E(a)^2 = a^2 - a^2 = 0

Use this property, and the assumption of independence, to compute the variances, and hence the standard deviations:

V(X + 10) = V(X) ==> SD(X + 10) = SD(X) = $90

V(5Y) = 5^2 V(Y) = 25 V(Y) ==> SD(5Y) = 5 SD(Y) = $40

V(X + Y) = V(X) + V(Y) ==> SD(X + Y) = √[SD(X)^2 + SD(Y)^2] = √8164 ≈ $90.35

8 0

4 years ago