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egoroff_w [7]
3 years ago
7

Mr. Brown's statistics class consists of 21 seniors. The class took two quizzes in one week, and 12 of the students in the class

passed both quizzes. What is the probability that the same students who passed the first quiz also passed the second quiz given that 19 students passed the first quiz?​
Mathematics
2 answers:
NeX [460]3 years ago
8 0

Answer:

That would be 63%.

Step-by-step explanation:

Here it is Adam.

damaskus [11]3 years ago
4 0
There is a 63% chance they passed
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Please help! this is timed!
Scilla [17]

Answer: -8-10

Step-by-step explanation:

4 0
3 years ago
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The shape of the sign outside Bob's Burger Barn is a regular octagon. How many degrees are in the measure of an interior angle o
DochEvi [55]

Answer:

An octagon has eight interior angles.

Interior angle total sum = (n-2) x 180, where n = number of sides.

Since an octagon has 8 sides, we have:

6 x 180 / 8, which is 135.

The answer is 135.

Let me know if this helps!

7 0
3 years ago
Read 2 more answers
The answer to this word problem usinf discriminant
solniwko [45]
\bf h=-9.8t^2+45t+1.2\implies 64.5=-9.8t^2+45t+1.2
\\\\\\
0=-9.8t^2+45t-63.3\\\\
-------------------------------

\bf \qquad \qquad \qquad \textit{discriminant of a quadratic}
\\\\\\
0=\stackrel{\stackrel{a}{\downarrow }}{-9.8}t^2\stackrel{\stackrel{b}{\downarrow }}{+45}t\stackrel{\stackrel{c}{\downarrow }}{-63.3}
~~~~~~
\stackrel{discriminant}{b^2-4ac}=
\begin{cases}
0&\textit{one solution}\\
positive&\textit{two solutions}\\
negative&\textit{no solution}
\end{cases}
\\\\\\
(45)^2-4(-9.8)(-63.3)\implies -456.36

the discriminant is a negative value, thus no solution for such quadratic, meaning if we use h=64.5 like we did, there's no "t" seconds at which point the ball hits the batting cage.
8 0
3 years ago
A bag of Halloween M&M’s contains 4 orange, 5 brown and 6 yellow candies. Two candies are selected without replacement. Let
Montano1993 [528]
P(A)= 4+5+6= 15/2= 75%
P(B)= 6+4=10  5/10= 50%
P(C)= 5+4=9    6/9= 66.6%

4 0
3 years ago
The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes
Artist 52 [7]

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(completing the exam in one hour or less)

P(x < 60)

P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)

Calculation the value from standard normal z table, we have,  

P(x < 60) =0.023= 2.3\%

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%

c) P(completing the exam in more than 90 minutes)

P(x > 90)

P( x > 90) = P( z > \displaystyle\frac{90 -80}{10}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 90) = 1 - 0.8413 = 0.1587 = 15.87\%

15.87% of children of class will require more than 90 minutes to complete the test.

Number of children =

\dfrac{15.87}{100}\times 60 = 9.52\approx 10

Approximately, 10 students of class will require more than 90 minutes to complete the test.

4 0
3 years ago
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