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3 years ago

The data set shows how many points Mark scored in each basketball game he played this season.

Mathematics

1 answer:

algol [13]3 years ago

5 0

Answer:

Kindly check attached picture

Step-by-step explanation:

Given the data :

1,4,4,5,5,7,8,8,10,11,11,11,14,15,16

The median (Q2) :

0.5(n + 1)th value

n = number of observations = 15

0.5(15 + 1)th value = 0.5(16) = 8th observation = 8

Q3:

0.75(n+1)th value = 0.75(16) = 12th observation = 11

Q1 :

0.25(n+1)th value = 0.25(16) = 4th observation = 5

Interquartile range (IQR) = (Q3 - Q1) :

11 - 5 = 6

MINIMUM VALUE IN DATASET = 1

MAXIMUM VALUE IN DATASET = 16

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PLEASE HELP I WILL GIVE YOU A BRAINLIEST! During a basketball game, Tom attempted 25 shots and made 9. He says he made 40% of th

postnew [5]

Answer:

Correct I think

Step-by-step explanation:

9 is nearly half of 25. (12.5) Half is equal to 50%. 40 is close to 50 and 9 is close to 12.5. So, he is correct. I think, hope this helps.

7 0

3 years ago

3.) Duncan spent $70 and now has<br> $120. How much money did he have<br> before spending the 70?

frosja888 [35]

Answer:

He started with $190

7 0

3 years ago

Read 2 more answers

The endpoints of the diameter of a circle are (–3, –6) and (–7, 8). Find the x-coordinate of the center of the circle.

Crazy boy [7]

You need to find the midpoint of the line. To get x coordinate you take (x1 + x2)/2 which is (-3+-7)/2 = -10/2 = -5. Same thing applies for y.

7 0

3 years ago

Monica's Money

pentagon [3]

Answer: The y-intercept should be the $20. The equation would be y=10x+20. Not sure if that’s what you’re asking.

Step-by-step explanation:

5 0

2 years ago

Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100.

sergejj [24]

Answer:

a) the probability that the minimum of the three is between 75 and 90 is 0.00072

b) the probability that the second smallest of the three is between 75 and 90 is 0.396

Step-by-step explanation:

Given that;

fx(x) = { 1/5 ; 50 < x < 100

0, otherwise}

Fx(x) = { x-50 / 50 ; 50 < x < 100

1 ; x > 100

n = 3

F(1) (x) = nf(x) ( 1-F(x)^n-1

= 3 × 1/50 ( 1 - ((x-50)/50)²

= 3/50 (( 100 - x)/50)²

=3/50³ ( 100 - x)²

Therefore P ( 75 < (x) < 90) = ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx

= 3/50³ [ -2 (100 - x ]₇₅⁹⁰

= (3 ( -20 + 50)) / 50₃

= 9 / 12500 = 0.00072

f(k) (x) = nf(x) ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k

Now for n = 3, k = 2

f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))

= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)

= 6/50³ ( 150x - x² - 5000 )

therefore

P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx

= 99 / 250 = 0.396

3 0

2 years ago