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IceJOKER [234]
3 years ago
14

In all, 1000 students took a statistics exam worth 150 points. The first quartile (or Q1) for all 1000 scores is 90 points. Abou

t how many students had scores greater than 90 points?
Mathematics
2 answers:
nasty-shy [4]3 years ago
5 0

Answer:

i think 750

Step-by-step explanation:

Brums [2.3K]3 years ago
4 0

Answer:

The number of students who scored more than 90 points is 750.

Step-by-step explanation:

Quartiles are statistical measures that the divide the data into four groups.

The first quartile (Q₁) indicates that 25% of the observation are less than or equal to Q₁.

The second quartile (Q₂) indicates that 50% of the observation are less than or equal to Q₂.

The third quartile (Q₃) indicates that 75% of the observation are less than or equal to Q₃.

It is provided that the first quartile is at 90 points.

That is, P (X ≤ 90) = 0.25.

The probability that a student scores more than 90 points is:

P (X > 90) = 1 - P (X ≤ 90)

                = 1 - 0.25

                = 0.75

The number of students who scored more than 90 points is: 1000 × 0.75 = 750.

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M∠A = 75 + 5x
Firlakuza [10]

Answer:

See below

Step-by-step explanation:

For a parallelogram m < A = m < C so:-

75 + 5x = 87 + 3x

2x = 12

x = 6  

and m< A = 75 + 30 = 105

and m < C = 87 + 18 = 105

Now if the other 2 angles are  = 75 degrees( that is the same side angles add up to 180 degrees) we have proved that it is a parallelogram.

m < B = m < D

111 - 6x = 39 + 6x

-12x = -72

x = 6

m < B = 111 - 36 = 75 and m < D = 39 + 36 = 75

So this has been proved.

6 0
3 years ago
Round 4,398,202 to the nearest 100
Klio2033 [76]
In the hundreds spot is a 2 and 2 is under 5 so we round down which would be 4 398 200
5 0
3 years ago
Read 2 more answers
On average, Tanner has noticed that 22 trains pass by his house daily (24 hours) on the nearby train tracks. What is the probabi
blondinia [14]

Answer:

0.11069

Step-by-step explanation:

We will assume that the trains pass by his house following a uniform distribution with values between 0 and 24. The probability of a train passing on a 9-hour time period is 9/24 = 3/8 = 0.375. Lets call Y the amount of trains passing by his house during that 9-hour period. Y follows a Binomail distribution with parameters 22 and 0.375.

P(Y ≤ 5) = P(Y = 0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) =

{22 \choose 0} * 0.375^0*(1-0.375)^{22} + {22 \choose 1}*0.375^1*(1-0.375)^{21} +\\\\{22 \choose 2} * 0.375^2*(1-0.375)^{20} + {22 \choose 3}*0.375^3*(1-0.375)^{19}  + \\{22 \choose 4} * 0.375^4*(1-0.375)^{18} + {22 \choose 5}*0.375^5*(1-0.375)^{17} = 0.11069

I hope that works for you!

4 0
3 years ago
4 (4b + 3) + 2b = 3 (6 - 11)
erma4kov [3.2K]
Solve for b?
1. Distribute “4(4b+3)” and “3(6-11)”
16b+12+2b=18-33

2. combine like terms
18b+12=-15
18b=-27
b=-27/18

3. simplify
b=3/2
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2 years ago
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Getting at least a 80% ( B average) is also good still very close to 90
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