Take the 25 art projects devised by 100% would be .25 then multiply .25 by the percentage that got perfect scores in which you would get 3 perfect scores
.25x12=3
4x+2x=180
6x= 180
Divide by 6
X= 30
Complete question is:
Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 29.7 grams per pound. Find the probability of observing a sample mean of 29.7 grams of fat per pound or less in a random sample of 34 farm-raised trout. Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Answer:
Probability = 0.0277
Step-by-step explanation:
We are given;
Mean: μ = 32
Standard deviation;σ = 7
Random sample number; n = 34
To solve this question, we would use the equation z = (x - μ)/(σ/√n) to find the z value that corresponds to 29.7 grams of fat.
Thus;
z = (29.7 - 32)/(7/√34)
Thus, z = -2.3/1.200490096
z = -1.9159
From the standard z table and confirming with z-calculator, the probability is 0.0277
Thus, the probability to select 34 fish whose average grams of fat per pound is less than 29.7 = 0.0277
A hemisphere is half a sphere.
So, the volume of a hemisphere is the half of the volume of the complete sphere.
The formula for the volume of a sphere is:
V = (4/3) π (radius)^3
Using π = 3.14 and radius = diameter / 2 = 50 feet / 2 = 25 feet.
You get:
V = (4/3) (3.14) (25 feet)^3 = 65,416.7 feet^3
Answer: 65,416.7 cu. ft.
Answer:
Step-by-step explanation:
q is TFTF
~q use negation, not q so is the opposite of q : FTFT
p↔~q use biconditional ,and will be True only is both statements are T or both are F
p values are TTFF ↔~q values are FTFT : FTTF
(p↔~q )∧~q use conjunction, that is True only if both statements are T
(p↔~q ) values are FTTF ∧~q values are FTFT : FTFF
(p↔~q )∧~q → p use a conditional statement, where only True False will give a F
(p↔~q )∧~q values are FTFF → p values are TTFF : TTTT