All categories

3 years ago

Out of a group of 250 pupils, 10 pupils could not join the school's excursion due to illness.

Mathematics

1 answer:

Arada [10]3 years ago

8 0

Answer:

out of 250 wala nagka crush sayo hshshs

charot lng

You might be interested in

What is the decimal equivalent of 13/4?

Gelneren [198K]

Answer:

3.25

hope this helped!!

7 0

2 years ago

Read 2 more answers

I need to find the limit

Digiron [165]

For $x=\dfrac{1}{2}$ the function is undefined.

$\displaystyle
\lim_{x\to\tfrac{1}{2}^-} x\sec (\pi x)=\infty\\
\lim_{x\to\tfrac{1}{2}^+} x\sec (\pi x)=-\infty\\\\
\lim_{x\to\tfrac{1}{2}^-}x\sec (\pi x)\not = \lim_{x\to\tfrac{1}{2}^+}x\sec (\pi x)
$
So, the limit doesn't exist.

8 0

3 years ago

Lowest common factor of 6 and 12

xxTIMURxx [149]

Answer:

Step-by-step explanation:

It's so obvious lollllll

5 0

2 years ago

Read 2 more answers

Six Republicans and four Democrats have applied for two open positions on a planning committee. Since all the applicants are qua

Dafna11 [192]

Answer:

$\displaystyle P=\frac{7}{15}=0.467$

Step-by-step explanation:

<u>Probabilities</u>

When we choose from two different sets to form a new set of n elements, we use the so-called hypergeometric distribution. We'll use an easier and more simple approach by the use of logic.

We have 6 republicans and 4 democrats applying for two positions. Let's call R to a republican member and D to a democrat member. There are three possibilities to choose two people from the two sets: DD, DR, RR. Both republicans, both democrats and one of each. We are asked to compute the probability of both being from the same party, i.e. the probability is

$P=P(DD)+P(RR)$

Let's compute P(DD). Both democrats come from the 4 members available and it can be done in $\binom{4}{2}$ different ways.

For P(RR) we proceed in a similar way to get $\binom{6}{2}$ different ways.

The total ways to select both from the same party is

$\displaystyle \binom{4}{2}+\binom{6}{2}=4+15=21$

The selection can be done from the whole set of candidates in $\binom{10}{2}$ different ways, so

$\displaystyle P=\frac{21}{\binom{10}{2}}$

$\displaystyle P=\frac{21}{45}=\frac{7}{15}=0.467$

3 0

3 years ago

A charity fundraiser has collected $750 in donations and plans to sell magazine ads to exceed its goal of $1200. Each ad will pr

sashaice [31]

Hi!

The first thing I would do is see how much the charity has to make to exceed its goal. To do that, I would subtract how much they've already made from their goal.

1200 - 750 = 450

Now, since each ad provides 25 dollars, and the charity wants to exceed its goal, add 25 to 450, to get 475. (What they have to make to exceed their goal)

Now, just find how many ads must be made, or how many 25 dollars, go into 475, by dividing 475 by 25.

475/25 = 19 ads

Of course, if you get this question wrong and instead its 18 because your teacher stated they weren't supposed to exceed the goal and instead just meet it, you can question their wording of the problem.

Hope this helped!

6 0

3 years ago

Read 2 more answers