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4 years ago

I NEED HELP!! What is the interquartile range (IQR) of the data?

Mathematics

1 answer:

Tom [10]4 years ago

4 0

What grade are you in? I don't believe I have learned this yet!

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Replace A with a number to make A/24 ≥ 1/4 a true statement.

antiseptic1488 [7]

Answer:

B. 7

Step-by-step explanation: 6/24= 1/4 so 7/24>1/4

7 0

3 years ago

FIRST PERSON TO ANSWER CORRECTLY GET MARKED BRAINLY

Ilya [14]

Answer:

n = -2 .................

8 0

3 years ago

Pls help me, it is so proooooooooooooooooooooo

Semmy [17]

Answer:

Step-by-step explanation:

i think it is A

because

$\frac{x}{3} + 10 = \frac{15}{3} + 10 = 5 + 10 = 15$

hope it helps

have a nice day

6 0

3 years ago

Read 2 more answers

PT = 7x + 8 and TQ = 9x - 6

erastova [34]

Answer:

t is 7

Step-by-step explanation:

i did it

4 0

3 years ago

Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou

monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

B) 0.6796

C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = $P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )$

= P ( Z > 2.33 ) = 0.0099

= 0.9%

B) percentage of oranges weighing less than 8.7 ounces

P( x < 8.7 ) = $P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )$

= P ( Z < 0.4667 ) = 0.6796

= 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = $P (\frac{6.2-8}{1.5} < \frac{x - u}{ std} < \frac{7-8}{1.5} )$

= P ( -1.2 < Z < -0.66 )

= Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0

4 years ago