-10 + -v - 2 + -9
First, simplify. / Your problem should look like:
Second, simplify/subtract. / Your problem should look like:
Answer:
-v - 21
Answer:
Y intercept of this function is 2 which is the height of a plant at the time x = 0 or the height of a plant when kept in sunlight.
Step-by-step explanation:
In the given function y = (2 + 4x), x is the number of hours a plant kept in sunlight and y is the height in mm.
Since this equation is in the form of equation of straight line y = mx +c in which m denotes gradient of the line and c is y intercept.
In y = 4x + 2, y intercept of the equation is 2 which means when a plant was kept in sunlight its height was 2 mm.
Answer:
y=−x−1/3x−8
Step-by-step explanation:
y=8x−1/3x+1
To find the inverse function, swap x and y, and solve the resulting equation for x.
If the initial function is not one-to-one, then there will be more than one inverse.
So, swap the variables: y=8x−1/3x+1 becomes x=8y−1/3y+1.
Now, solve the equation x=8y−1/3y+1 for y.
y=−x−1/3x−8
Sorry I was in a rush and couldn't do the fractions as formulae. Hope it helped anyways.
Answer:
a)0.099834
b) 0
Step-by-step explanation:
To solve for this question we would be using , z.score formula.
The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.
a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.
Standard Deviation = √variance
= √0.16 = 0.4
Standard deviation = 0.4
Mean = 21.37
x = 20.857
z = (x-μ)/σ
z = 20.857 - 21.37/0.4
z = -1.2825
P-value from Z-Table:
P(x<20.857) = 0.099834
b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.
z score formula used = (x-μ)/σ/√n
x = 20.857
Standard deviation = 0.4
Mean = 21.37
n = 100
z = 20.857 - 21.37/0.4/√100
= 20.857 - 21.37/ 0.4/10
= 20.857 - 21.37/ 0.04
= -12.825
P-value from Z-Table:
P(x<20.857) = 0
c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.
Answer: 59 1/2
Step-by-step explanation:
