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2 years ago

10

There are 196 countries in the world. How many students must be enrolled in a university to guarantee that there are at least 20

0 who come from the same country?

1 answer:

cestrela7 [59]2 years ago

3 0

Answer:

At least 39200 students.

Step-by-step explanation:

Given that there are 196 countries in the world, and each country would have at least 200 students in a university.

For each country to have at least 200 students in the university, then;

Number of students enrolled in the university = number of required students x number of countries

= 200 x 196

= 39200

At least, 39200 students must be enrolled in the university. Provided that the admission procedures are conditioned for the purpose.

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A bakery offers a sale price of $2.75 for 3 muffins. What is the price per dozen

Svet_ta [14]

Answer:

4 muffins is $2.75

4*3=12

1 dozen is 12

2.75 * 3 =$ 8.25

Step-by-step explanation:

6 0

3 years ago

Please help me solve the table below

ArbitrLikvidat [17]

Answer:

2, 4 ,6 8

Step-by-step explanation:

Given condition y = 2x

When x = 1 , y = 2 *1 = 2

When x = 2 , y = 2 * 2= 4

when x = 3 , y = 2* 3 = 6

When x = 4 , y = 2 * 4 = 8

Hope it will help :)

5 0

2 years ago

If PN=29cm and MN=13 , then PM=?

HACTEHA [7]

*see attachment for the figure referred to

Answer/Step-by-step explanation:

1. PN = 29

MN = 13

PM = ?

$PN = PM + MN$ (Segment addition postulate)

$PN - MN = PM$ (subtract MN from each side)

$29 - 13 = PM$ (substitute)

$16 = PM$

$PM = 16 cm$

2. PN = 34, MN = 19, PM = ?

$PN = PM + MN$ (sediment addition postulate)

$PN - MN = PM$ (subtract MN from each side)

$34 - 19 = PM$ (substitute)

$15 = PM$

$PM = 15 cm$

3. PM = 19, MN = 23, PN = ?

$PN = PM + MN$ (Segment addition postulate)

$PN = 19 + 23$ (substitute)

$PN = 42 cm$

4. MN = 82, PN = 105, PM = ?

$PN = PM + MN$ (segment addition postulate)

$PN - MN = PM$ (subtract MN from each side)

$105 - 82 = PM$ (substitute)

$23 = PM$

$PM = 23 cm$

5. PM = 100, MN = 100, PN = ?

$PN = PM + MN$ (Segment addition postulate)

$PN = 100 + 100$ (substitute)

$PN = 200 cm$

7 0

3 years ago

What is the amplitude of sin ?

sp2606 [1]

You haven't provided a graph or equation so I will tell the simplified meaning of amplitude instead.

Amplitude, is basically a distance from midline/baseline to the maximum or minimum point.

For sine function, can be written as:

$\displaystyle \large{ y = A \sin(bx - c) + d}$

A = amplitude
b = period = 2π/b
c = horizontal shift
d = vertical shift

I am not able to provide an attachment for an easy view but I will try my best!

We know that amplitude or A is a distance from baseline/midline to the max-min point.

Let's see the example of equation:

$\displaystyle \large{y = 2 \sin x}$

Refer to the equation above:

Amplitude = 2
b = 1 and therefore, period = 2π/1 = 2π
c = 0
d = 0

Thus, the baseline or midline is y = 0 or x-axis.

You can also plot the graph on desmos, y = 2sinx and you will see that the sine graph has max points at 2 and min points at = -2. They are amplitude.

So to conclude or say this:

If Amplitude = A from y = Asin(x), then the range of function will always be -A ≤ y ≤ A and have max points at A; min points at -A.

6 0

2 years ago

The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =

gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy is given as :

$f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0$

Thus, the expected total claim amount $\mu$ = 1000

The variance of the total claim amount $\sigma ^2 = 1000^2$

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

$P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})$

$P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})$

$P(X>1100n) = P(Z> \dfrac{10*100}{1000})$

$P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345$

$\mathbf{P(X>1100n) = 0.158655}$

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

4 0

2 years ago