A soft drink machine outputs a mean of 27 ounces per cup. The machine's output is normally distributed with a standard deviation of 3 ounces. What is the probability of filling a cup between 22 and 29 ounces? Round your answer to four decimal places.

P(22 ≤ x ≤ 29) = 0.703

Step-by-step explanation:

Since the machine's output is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = output of the machine in ounces per cup.

µ = mean output

σ = standard deviation

From the information given,

µ = 27

σ = 3

The probability of filling a cup between 22 and 29 ounces is expressed as

P(22 ≤ x ≤ 29)

For x = 22,

z = (22 - 27)/3 = - 1.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.047

For x = 29,

z = (29 - 27)/3 = 0.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.75

Therefore,

P(22 ≤ x ≤ 29) = 0.75 - 0.047 = 0.703

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°