What’s is two plus the quotient of a number and five is the same as ten? As an equation?

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b

horrorfan [7]

Answer:

a. $X\sim N(\mu = 6.1, \sigma = 1.9)$ b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is $z_{1} = (6.4-6.1)/1.9 = 0.1579$ and the z-score related to 7 kg is $z_{2} = (7-6.1)/1.9 = 0.4737$ , we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0

3 years ago

A clothes inspector examines 75 randomly chosen garments a day. If 10 of them have a defect, about how many garments are defecti

Nookie1986 [14]

950 garments would be defective

8 0

3 years ago

Kendra used these steps to solve the equation 2x + 5 = 3 + 2(x + 1). Which choice describes the meaning of her result, 5 = 5?

alekssr [168]

The answer is A.
If a redundant conclusion is reached in basic algebra this states that the variable holds all possible real values.
If you algebraically solve Kendra's you do achieve the true statement 5 = 5 (leaving out D). And if you test any value of x for the equation it does hold true (getting rid of B).
Hopefully this makes sense.

6 0

4 years ago

Read 2 more answers

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

The magnitude and direction of two forces acting on an object are 100100 pounds, S7878degrees°E, and 5050 pounds, N5353degree

amm1812

Answer:

138.06 lb N86.1°E

Step-by-step explanation:

There are several ways you can work this. One of the most straightforward is to resolve each vector into its north and east components, add those, and then covert the result back to magnitude and direction.

A diagram can help immensely.

We note that S78°E is the same as E12°S. Since we're used to seeing the coordinate system with the +x axis aligned with east, it can be convenient to think of the first force as 100 lb at -12°.

The angle of the second force is N53°E, which can be expressed as E37°N. Then in x-y coordinates, this force is 50 lb at +37°.

The components of the sum are the sum of the components:

R = F1 +F2 = (100cos(-12°) +50cos(37°), 100sin(-12°) +50sin(37°))

= (137.747, 9.300)

The the magnitude of the resultant is computed using the Pythagorean theorem:

||R|| = √(137.747² +9.300²) ≈ 138.06

and the angle is computed using the arctangent function. Here, our diagram tells us the angle is in the first quadrant, so is positive (relative to +x, or East).

∠R = arctan(9.300/137.747) ≈ 3.9°

__

We want to express the answer in terms similar to the way the given forces are expressed, so we want the angle relative to north. The resultant is then described by ...

R = 138.06 lb at N86.1°E

4 0

4 years ago