Two numbers differ by 13 and the sum of their squares is 125 find the numbers

F(x)=4x-1 G(x)=x+3 Find f(x)+g(x)

mr_godi [17]

Answer:

Answer is 5x+2

Step-by-step explanation:

$Given \: f(x) = 4x - 1 \: \: and \: \\ g(x) = x + 3 \\ f(x) + g(x) = 4x -1 + x + 3 \\ = 5x + 2$

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8 0

3 years ago

The coordinates of the vertices of a rectangle are (-2, 3), (4, -4), (4, 3), and (-2, -4). What are the dimensions of the rectan

erik [133]

Answer:

D

Step-by-step explanation:

Graph the points. Then count the units vertically and horizontally. See picture.

7 0

3 years ago

Karen is working on her math homework. She solves the equation

Gelneren [198K]

Disagree.
b/8 = 56; multiply both sides by 8 to solve for b, and you get b = 448

5 0

3 years ago

A rack has a volume of 10 cm3 and a mass of 25g. What is the density?

valentinak56 [21]

Answer:

Step-by-step explanation:

density is mass divided by volume, so 25g divided by 10 cm = 2.5 g

4 0

3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

(c) $P(X > 980) = 0.00001\\\\$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

3 years ago