If a quadrilateral is a rhombus, what must also be true?

I need help??? f(x) = 3x + 10x and g(x) = 4x – 2, find (f – g)(x)

goldfiish [28.3K]

F(x) = 3ˣ + 10x and g(x) = 4x – 2, find (f-g)(x)
(f-g)(x) = f(x) - g(x):

(3ˣ + 10x) - (4x - 2) →→→ 3ˣ + 6x - 2. Then:

(f-g)(x) = 3ˣ + 6x - 2

5 0

4 years ago

HELP ALERT ALERT MATH PROBLEM

Lisa [10]

Multiple 15 by 6, the answer is 90 square cm

7 0

3 years ago

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an

Gelneren [198K]

Answer:

(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091

(c) P (Z < -1.22) = 0.1112 (d) P (1.13 < Z > 3.35) = 0.1288

(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, $X \sim N (\mu, \sigma^{2})$ , then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

= 1 - 0.99086

= 0.00914

≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

= 0.99960 - 0.87076

= 0.12884

≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35) is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

= 0.29116 - 0.22065

= 0.07051

≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55) is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

= 1 - 0.99865

= 0.00135

≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

= 0.99948

≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

3 0

3 years ago

Solve for x 3(8x-5)=-4(7-6x)

tekilochka [14]

3(8x - 5) = -4(7 - 6x)

Distribute 3 and -4 inside their respective parentheses.

24x - 15 = -28 + 24x

Add 28 to both sides.

24x + 13 = 24x

Subtract 24x from both sides.

13 = 0

This is an untrue statement, so that means there are no solutions to this equation.

Your answer is no solution.

6 0

3 years ago

Read 2 more answers

Use the chain rule to find ∂z/∂s and ∂z/∂t. z = sin(θ) cos(ϕ), θ = st5, ϕ = s7t

8090 [49]

$z=\sin\theta\cos\phi$
$\theta=st^5$
$\phi=s^7t$

$\dfrac{\partial z}{\partial s}=\dfrac{\partial z}{\partial\theta}\dfrac{\partial\theta}{\partial s}+\dfrac{\partial z}{\partial\phi}\dfrac{\partial\phi}{\partial s}$
$\dfrac{\partial z}{\partial s}=\cos\theta\cos\phi t^5-\sin\theta\sin\phi (7s^6t)$
$\dfrac{\partial z}{\partial s}=t^5\cos(st^5)\cos(s^7t)-7s^6t\sin(st^5)\sin(s^7t)$

$\dfrac{\partial z}{\partial t}=\dfrac{\partial 
z}{\partial\theta}\dfrac{\partial\theta}{\partial t}+\dfrac{\partial 
z}{\partial\phi}\dfrac{\partial\phi}{\partial t}$
$\dfrac{\partial z}{\partial t}=\cos\theta\cos\phi(s5t^4)-\sin\theta\sin\phi s^7$
$\dfrac{\partial z}{\partial t}=5st^4\cos(st^5)\cos(s^7t)-s^7\sin(st^5)\sin(s^7t)$

7 0

3 years ago