Answer:
Step-by-step explanation:
Since the results for the standardized test are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = test reults
µ = mean score
σ = standard deviation
From the information given,
µ = 1700 points
σ = 75 points
We want to the probability that a student will score more than 1700 points. This is expressed as
P(x > 1700) = 1 - P(x ≤ 1700)
For x = 1700,
z = (1700 - 1700)/75 = 0/75 = 0
Looking at the normal distribution table, the probability corresponding to the z score is 0.5
P(x > 1700) = 1 - 0.5 = 0.5
Step-by-step explanation:
a little messy but hope it helps :)) !!
Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:

In which:
is the sample proportion.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
The sample size and the estimate are given by:

Hence:


The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).
More can be learned about the z-distribution at brainly.com/question/25890103
For this case what you should do is follow the following steps:
1) factorize numerator and denominator of both expressions.
2) cancel similar terms
3) add both fractions by cross product
4) Rewrite the numerator and denominator.
Answer:
See attached image.
The value of the "6" in 49.62 is in the<u> tenth</u> place