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zheka24 [161]
3 years ago
7

Class records at rockwood college indicate that a student selected at random has probability 0.75 of passing french 101. for the

student who passes french 101, the probability is 0.91 that he or she will pass french 102. what is the probability that a student selected at random will pass both french 101 and french 102? (round your answer to three decimal places.)
Mathematics
2 answers:
lilavasa [31]3 years ago
7 0
Please Send Lions, Monkeys, Cats And Zebras Into Lovely Hot Countries
Signed General Penguin
P - potassium
S - sodium
L - lithium
M - magnesium
C - calcium
A - aluminium
Z - zinc
I - iron
L - lead
H - hydrogen
C - copper
S - silver
G - gold
<span>P - platinum</span>
Kipish [7]3 years ago
4 0

To get the probability of two individual events both occurring, you have to multiply the probabilities of their individual events occurring. Therefore in this problem, the probability that a student selected at random will pass both French 101 and French 102 is 0.683 (.75 x .91). The answer is already rounded to three decimal places.

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Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar
klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack

Where (from tables):

Z_{0.99}=2.33

Finally, the interval at 98% confidence level is:

CI(\mu)=\lbrack28.94,31.06\rbrack

4 0
1 year ago
What is the answer??
masha68 [24]
The answer is 30, assuming it is a cube, 
7 0
3 years ago
Try this trick out on a friend. Tell your friend to place a dime in one hand and a penny in the other hand. Explain that you can
lyudmila [28]

Answer:

does it work

Step-by-step explanation:

7 0
3 years ago
What variable expressions is a translation of the word phrase "three less than the product of a number and six?
yan [13]

That expression would be 6n-3, where n is the number in question.


6 0
3 years ago
Write a recursive rule and an explicit rule for the<br> sequence<br> 3,7,11,15
m_a_m_a [10]

Answer:

The Recursive formula for the sequence is:

aₙ = aₙ₋₁ + d

The Explicit formula for the sequence is:

a_n=4n-1

Step-by-step explanation:

Given the sequence

3,7,11,15

Here:

a₁ = 3

computing the differences of all the adjacent terms

7 - 3 = 4, 11 - 7 = 4, 15 - 11 = 4

The difference between all the adjacent terms is the same and equal to

d = 4

We know that a recursive formula basically defines each term of a sequence using the previous term(s).

The recursive formula of the Arithmetic sequence always involves the first term.

a₁ = 3

We know that, in the Arithmetic sequence, every next term can be obtained by adding the common difference and the preceding term.

so

The recursive formula of the sequence is:

aₙ = aₙ₋₁ + d

substitute n = 2 to find the 2nd term

a₂ = a₂₋₁ + d

a₂ = a₁+ d

substitute a₁ = 3 and d = 4

a₂ = 3 + 4

a₂ = 7

Thus, the recursive formula for the sequence 3,7,11,15 is:

aₙ = aₙ₋₁ + d

<u>An explicit rule for the sequence</u>

Given the sequence

3,7,11,15

We already know that

a₁ = 3

d = 4

An arithmetic sequence has a constant difference 'd' and is defined by  

a_n=a_1+\left(n-1\right)d

substituting a₁ = 3 and d = 4

a_n=4\left(n-1\right)+3

a_n=4n-4+3

a_n=4n-1

Therefore, an explicit rule for the  sequence

a_n=4n-1

6 0
3 years ago
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