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Marina CMI [18]
2 years ago
8

I need someone to solve this equation

Mathematics
1 answer:
vaieri [72.5K]2 years ago
7 0
I’m pretty sure it’s x=C-6
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A basketball player has a 50% chance of making each free-throw. What is the probability that the player makes at least 11 out of
vitfil [10]

Answer:

100/2048=0.048828125%

Step-by-step explanation:

He has a 50% chance of making each free-throw, so 1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2=1/(2^11)=1/2048

to get a percentage you time by 100 to get 100/2048

8 0
3 years ago
The average height of a 13 year old male in the U.S. is 60 inches, with a standard deviation of 2 inches. The average weight of
Romashka [77]

Since the average height is 60 inches and its deviation is 2 inches, one deviation to the right (or higher) is 62 inches (60 + 2). Two deviations is 64 inches, three deviations is 66 inches, and four deviations is 68 inches.


Since the average weight is 100 pounds and its deviation is 5 inches, we repeat the process from finding heights to get to 115 pounds. That takes three deviations.


The MORE deviations away, the more unusual it is. So the height (4 deviations) is more unusual than the weight (3 deviations).

5 0
3 years ago
Read 2 more answers
Help on these problems please for my test ​
kvv77 [185]

Answer:

120

Step-by-step explanation:

5 0
3 years ago
Which linear inequality represents the graph below?
musickatia [10]

Answer:

A.

Step-by-step explanation:

It has a negative slope, so B and C are eliminated. The graph also has a solid line instead of a dashed line. This means D is eliminated because it is only greater than. If the graph had a dashed line, then D would be correct, but it does not.

5 0
3 years ago
Find the missing dimension. Use the scale factor 1: 15. Door Model Height 10 in Actual
anygoal [31]

Answer:

The answer is "150 unit"

Step-by-step explanation:

scale factor = 1:15

height = 10

dimension =?

let dimension = x

\Rightarrow \frac{10}{x} = \frac{1}{15} \\\\\Rightarrow  150 = x\\\\\Rightarrow x= 150 \ units

5 0
3 years ago
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