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3 years ago

A basketball player Made 63 out of 75 free throws what percent is this

Mathematics

1 answer:

Kazeer [188]3 years ago

3 0

Answer: 84%

Step-by-step explanation:

1. You divide: 63/75

2. Your answer is 0.84

3. You convert it into a percent by moving the decimal two times to the right

4. You get 84%

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How many times does 16 go into 40

qwelly [4]

2.5 times.

40 divided by16 equals 2.5
Hope that helped!

8 0

3 years ago

Read 2 more answers

A researcher wishes to estimate the proportion of X-ray machines that malfunction. A random 275 sample of machines is taken, and

Dominik [7]

Answer:

Following are the solution to the given question:

Step-by-step explanation:

$95\%$ Confidence Interval for both the percentage of all x-ray machines

p = the machinery's share is not working:

$= \frac{228}{275}\\\\ = 0.829$

$\text{Margin of Error} = Z_{(\frac{\alpha}{2})} \times \sqrt{( p \times (1-p)}{n})$

$= 1.96 \times \sqrt{(0.829 \times \frac{0.171}{275})} \\\\= 1.96 \times 0.023 \\\\= 0.045$

Lower $95\%$ Confidence interval = p - error margin $= 0.829 - 0.045 = 0.784$

Upper $95\%$ Confidence Interval = p + error margin $= 0.829 + 0.045 = 0.874$

So, $95\%$ Confidence Interval $= ( 0.78 , 0.87 )$

4 0

3 years ago

12/65 into a simplified fraction

yanalaym [24]

12/65 divide by two if is simplified tell me ok

5 0

3 years ago

Read 2 more answers

Pls help. Take a pic of your work on paper thanks.

artcher [175]

Answers:

1.) -x - 5

2.) 11x^2 + 5x

3.) 13x + 2y +3y^2

4.) -12x - 2

5.) -24

Sorry, it won’t somehow let me attach a picture of my work on paper :(

But hopefully this helps! :)

6 0

3 years ago

Read 2 more answers

1.How many combinations are possible from the letters HOLIDAY if the letters are taken?

Lesechka [4]

Using the combination formula, it is found that:

1. A. 7 combinations are possible.

B. 21 combinations are possible.

C. 1 combination is possible.

2. There are 245 ways to group them.

<h3>What is the combination formula?</h3>

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Exercise 1, item a:

One letter from a set of 7, hence:

$C_{7,1} = \frac{7!}{1!6!} = 7$

7 combinations are possible.

Item b:

Two letters from a set of 7, hence:

$C_{7,2} = \frac{7!}{2!5!} = 21$

21 combinations are possible.

Item c:

7 letters from a set of 7, hence:

$C_{7,7} = \frac{7!}{0!7!} = 1$

1 combination is possible.

Question 2:

Three singers are taken from a set of 7, and four dances from a set of 10, hence:

$T = C_{7,3}C_{10,4} = \frac{7!}{3!4!} \times \frac{10!}{4!6!} = 245$

There are 245 ways to group them.

More can be learned about the combination formula at brainly.com/question/25821700

6 0

2 years ago