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2 years ago

What are 3(or less) conclusions that you can make based on this histogram?

Mathematics

1 answer:

jeyben [28]2 years ago

7 0

Answer:

1. theres way more younger people compared to older

2. theres a total of approximately 32 people ages 10-14

3. theres a total of approximately 31 people ages 14-19

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Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit 10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

Answer:

40.13%

Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is $\overline A=\textrm{Missing a target at least once}$

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

Now, $P(A)=0.95^{10}=0.5987$

We know that the sum of probability of an event and its complement is 1.

So, $P(A)+P(\overline A)=1\\\\P(\overline A)=1-P(A)\\\\P(\overline A)=1-0.5987\\\\P(\overline A)=0.4013=40.13\%$

Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.

6 0

3 years ago

Read 2 more answers

1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$

3 0

3 years ago

Differentiate the function with respect to x<br> f(x)= x^4/ x^2- 4<br> Show work

jeyben [28]

Answer:

Step-by-step explanation:

$f(x)=\frac{u}{v} ,u~ and~ v~ functions~ of~ x\\f'(x)=\frac{v ~u'-u~v' }{v^2} \\f(x)=\frac{x^4}{x^2-4} \\f'(x)=\frac{(x^2-4)*4x^3-x^4(2x)}{(x^2-4)^2} \\=\frac{2x^3(2x^2-8-x^2) }{(x^2-4)^2} \\=\frac{2x^3(x^2-8)}{(x^2-4)^2}$

7 0

2 years ago

Kinda want to cry right now.. Someone PLEASE help me.. I know there is someone out there smart enough.... 25 points and Brainlie

Svetlanka [38]

Answer:

a) Infinite solutions

b) No solutions

Step-by-step explanation:

First, know the following:

If the graphs intersect, there's only one solution.

If the graphs are parallel, there are no solutions.

If the graphs are the exact same line, there are infinite solutions.

For a):

Change the first equation into a linear one.

2x+3y=6

3y=-2x+6

$y=-\frac{2}{3} x+2$

Change the second equation into a linear one.

4x+6y=12
6y=-4x+12
$y=-\frac{2}{3} x+2$

Boom. You have two equations which are equal. As stated above, graphs on the exact same line have infinite solutions.

For b)

They are already in linear form so hurray.

$y=\frac{1}{5} x+2\\y=\frac{1}{5} x+10\\$

These lines are parallel since they have the SAME slope but a different y-intercept. As stated above, parallel lines have no solutions.

5 0

3 years ago

Find the Value of a and b i dare u

fenix001 [56]

Well it's a dare so I guess I'll do it.

3 0

3 years ago