X/y
- 8/20=42/y
- 0.4=42/y
- y=42/0.4
- y=105
A sequence<span> is a set of numbers.
</span>If the difference between two consecutive terms is constant, then the sequence is called arithmetic and th<span>e difference is called the common difference.
</span><span>If the numbers in the sequence follow a pattern were the next term is found by multiplying by a constant called the common ratio, r, then the sequence is geometric sequence.</span>
Answer:
The approximate percentage of SAT scores that are less than 865 is 16%.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
Approximately 68% of the measures are within 1 standard deviation of the mean.
Approximately 95% of the measures are within 2 standard deviations of the mean.
Approximately 99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean of 1060, standard deviation of 195.
Empirical Rule to estimate the approximate percentage of SAT scores that are less than 865.
865 = 1060 - 195
So 865 is one standard deviation below the mean.
Approximately 68% of the measures are within 1 standard deviation of the mean, so approximately 100 - 68 = 32% are more than 1 standard deviation from the mean. The normal distribution is symmetric, which means that approximately 32/2 = 16% are more than 1 standard deviation below the mean and approximately 16% are more than 1 standard deviation above the mean. So
The approximate percentage of SAT scores that are less than 865 is 16%.
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!
Answer:
7
Step-by-step explanation:
Changing LHS as
Comparing LHS and RHS
