The scores on a certain test are normally distributed with a mean score of 69 and a standard deviation of 4. what is the probability that a sample of 90 students will have a mean score of at least 69.4216? your answer should be a decimal rounded to the fourth decimal place.

Solution: We are given:

μ=69,σ=4,n=90

We have to find P(Mean≥69.4216)

We need to first find the z score

z= (xbar-μ)/(σ/sqrt(n))

=(69.4216-69)/(4/sqrt(90))

=1.00

Now we have to find P(z≥1.00)

Using the standard normal table, we have:

P(z≥1.00)= 0.1587

Therefore the probability that a sample of 90 students will have a mean score of at least 69.4216 is 0.1587.

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116