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amm1812
3 years ago
15

Please help! Correct answer only please! I need to finish this assignment by today.

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

1/2

Step-by-step explanation:

Since there are a total of 14 t shirts and 7 of them were green, the experimental probability that the next one is also green is 7/14=1/2. Hope this  helps!

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15% of R560 - 15% of R500 is:​
Dahasolnce [82]

15% of R560 - 15% of R500

=> 0.15 × 560 – 0.15 × 500

=> 0.15 ( 560–500)

=> 0.15 × 60

=> Rs. 9

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️

8 0
3 years ago
Explain the special pattern for sketching a parabola
Lady bird [3.3K]

Answer:

the parabola can be written as:

f(x) = y = a*x^2 + b*x + c

first step.

find the vertex at:

x  = -b/2a

the vertex will be the point (-b/2a, f(-b/2a))

now, if a is positive, then the arms of the parabola go up, if a is negative, the arms of the parabola go down.

The next step is to see if we have real roots by using the Bhaskara's equation:

x =  \frac{-b +-\sqrt{b^2 -4ac} }{2a}

Now, draw the vertex, after that draw the values of the roots in the x-axis, and now conect the points with the general draw of the parabola.

If you do not have any real roots, you can feed into the parabola some different values of x around the vertex

for example at:

x =  (-b/2a) + 1 and x =  (-b/2a) - 1

those two values should give the same value of y, and now you can connect the vertex with those two points.

If you want a more exact drawing, you can add more points (like x =  (-b/2a) + 3 and x =  (-b/2a) - 3) and connect them, as more points you add, the best sketch you will have.

8 0
3 years ago
The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo
muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: {\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}

Where \lambda is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}

Substitute the respective values in the above formula.

P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})

P(k > 2)= 1 - (0.44596)

P(k > 2)=0.55404

Hence, the required probability is 0.55404.

4 0
3 years ago
The graph of f(x) = is shown with g(x). Which equation represents the graph of g(x)? g(x) = g(x) = g(x) = + 1 g(x) = –1
tresset_1 [31]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
on a large college campus, 84% of the students report drinking alcohol within the past month, 33% report using some type of toba
Natasha_Volkova [10]

Answer:  0.31

Step-by-step explanation:

Let A denotes the event that the students report drinking alcohol and B denotes the students report using some type of tobacco product .

Given : P(A) =0.84   ; P(B)=0.33    and        P(A∪B)=0.86

We know that P(A\cap B)=P(A)+P(B)-P(A\cup B)

Then, the probability that the student both drunk alcohol and used tobacco in the past month is given by :-

 P(A\cap B)=0.84+0.33-0.86=0.31

Hence, the probability that the student both drunk alcohol and used tobacco in the past month = 0.31

5 0
3 years ago
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