Answer: The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is 0.596
Step-by-step explanation:
Since the weights of catfish are assumed to be normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = weights of catfish.
µ = mean weight
σ = standard deviation
From the information given,
µ = 3.2 pounds
σ = 0.8 pound
The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is is expressed as
P(x ≤ 3 ≤ 5.4)
For x = 3
z = (3 - 3.2)/0.8 = - 0.25
Looking at the normal distribution table, the probability corresponding to the z score is 0.401
For x = 5.4
z = (5.4 - 3.2)/0.8 = 2.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.997
Therefore,.
P(x ≤ 3 ≤ 5.4) = 0.997 - 0.401 = 0.596
Answer:
77.98
Step-by-step explanation:
You have two 30-60-90 triangles, ADC and BDC.
The ratio of the lengths of the sides of a 30-60-90 triangle is
short leg : long leg : hypotenuse
1 : sqrt(3) : 2
Using triangle ADC, we can find length AC.
Using triangle BDC, we can find length BC.
Then AB = AC - BC
First, we find length AC.
Look at triangle ACD.
DC is the short leg opposite the 30-deg angle.
DC = 10sqrt(3)
AC = sqrt(3) * 10sqrt(3) = 3 * 10 = 30
Now, we find length BC.
Look at triangle BCD.
For triangle BCD, the long leg is DC and the short leg is BC.
BC = 10sqrt(3)/sqrt(3) = 10
AB = AC - BC = 30 - 10 = 20
Answer: 1.8 bags
Step-by-step explanation:
From the question, given that robby have 4 4/9 bags of pet food, 2/5 were dog food
2/5 of 4 4/9 = dog foods
Convert 4 4/9 to proper fraction= 40/9
2/5 of 40/9 means 2/5 × 40/9
= 16/9
= 1.78 or 1.8
I hope this helps, please mark as brainliest answer.
Answer:Graph the image of the given triangle under a dilation with a scale factor of 12 and center of dilation (0, 0)
