9,500 just add the 0 always add 0 if you multiply by 10. 00 for 100 and 000 for 1000
Answer:
(3x + 10) / (2x + 5)(x - 5).
Step-by-step explanation:
(3/2x+5) + (5/x-5)
= [3(x - 5) + 5(2x + 5) ] / [ (2x + 5)(x - 5)]
= 3x - 15 + 10x + 25 / (2x + 5)(x - 5)
= 13x + 10 / (2x + 5)(x - 5).
Answer: $0.66
Step-by-step explanation: Just divide 3 / $1.98 and you will get $0.66.
Fairly simple, But the only thing I don't understand is why a melon is on sale for $0.66. Thats CHEAP! Give me 100 :)
Answer:
The minimum score needed to be considered for admission to Stanfords graduate school is 328.48.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the minimum score needed to be considered for admission to Stanfords graduate school?
Top 2.5%.
So X when Z has a pvalue of 1-0.025 = 0.975. So X when Z = 1.96




The minimum score needed to be considered for admission to Stanfords graduate school is 328.48.
Answer:
a) For the 90% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
b) For the 99% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
Step-by-step explanation:
Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
Solution to the problem
Part a
For the 90% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
Part b
For the 99% confidence interval the value of
and
, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got: