Please help me answer number 1

Consider the following problem: a box with an open top is to be constructed from a square piece of cardboard, 3 feet wide, by cu

Alecsey [184]

Answer:

) See annex

b) See annex

x = 0,5 ft

y = 2 ft and

V = 2 ft³

Step-by-step explanation: See annex

c) V = y*y*x

d-1) y = 3 - 2x

d-2) V = (3-2x)* ( 3-2x)* x ⇒ V = (3-2x)²*x

V(x) =( 9 + 4x² - 12x )*x ⇒ V(x) = 9x + 4x³ - 12x²

Taking derivatives

V¨(x) = 9 + 12x² - 24x

V¨(x) = 0 ⇒ 12x² -24x +9 = 0 ⇒ 4x² - 8x + 3 = 0

Solving for x (second degree equation)

x =[ -b ± √b²- 4ac ] / 2a

we get x₁ = 1,5 and x₂ = 0,5

We look at y = 3 - 2x and see that the value x₂ is the only valid root

then

x = 0,5 ft

y = 2 ft and

V = 0,5*2*2

V = 2 ft³

7 0

2 years ago

Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr

bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .