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zavuch27 [327]
3 years ago
13

A pro baseball team is having a promotion in which every 10th fan that enters the stadium gets a free hat and every 12th person

gets a free T-shirt .
How many fans Will come into the stadium before a fan receives both a hat and a T- shirt ?
Mathematics
1 answer:
mixas84 [53]3 years ago
3 0
Multiples of 10 = 10, 20, 30, 40, 50, 60
This means the 10th, 20th, 30th, 40th, 50th, 60th, ... fan will get a free hat.
Multiples of 12 = 12, 24, 36, 48, 60
This means the 12th, 24th, 36th, 48th, 60th, ... fan will get a free shirt.
So the 60th fan will receive both a hat and a shirt.
Meaning 60 fans will come to the stadium before a fan receives both a hat and a shirt.
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My grade is a 77%, if i get a 0% on a paper worth 30% of my grade, what will my grade be now??
blsea [12.9K]
Grade=Paper+Other
Grade=(Grade_{paper}\times Value_{paper})+(Grade_{other}\times Value_{other})
Grade=(0 \times 30\%)+(77\% \times (1-30\%))
Grade=(0 \times 30\%)+(77\% \times 70\%)
Grade=0+53.9\%
Grade=53.9\%
6 0
3 years ago
Identify the two triangle congruence criteria that do NOT guarantee congruence. Explain why they do not guarantee congruence, an
Mashcka [7]

Answer: Hello mate!

there are two congruence criteria that do not guarantee congruence:

AAA ( or 3 angles)

two triangles can have the same 3 angles, but different size, then they are not congruent; an example of this is:

take the triangle rectangle of both cathetus = 1 and hypotenuse = √2, where the angles are 90°, 45° and 45°

and now take the triangle rectangle with both cathetus = 2 and hypotenuse = √8, this triangle also has the angles 90°, 45°, and 45°, so this two triangles succeed the AAA criteria, but are not congruent.

SSA (side-side-angle)

If two triangles satisfy the SSA  criteria and the corresponding angles are acute and the length of the side opposite to the angle is greater than the length of the adjacent side multiplied by the sine of the angle (but less than the length of the adjacent side), then the two triangles are not necessarily congruent.

This is kinda harder to illustrate;  

think on a triangle rectangle where you have the measure of both cathetus and one of the angles different from 90° as the SSA data.

and now think on another triangle that has the same adjacent cathetus and angle, and where the other cathetus is rotated (in a sense where 90° is decreasing) to the point where its tip intercepts the hypotenuse of the first triangle.

Those two triangles meet the SSA criteria but are not congruent.

7 0
3 years ago
How many flyers can we buy?
Rudiy27

Answer: 1760

Step-by-step explanation:

Cost of 1 flyer = $0.25 or 25 cents

Budget is = $555

Let us assume that supplies cost is a one time cost

Cost of supplies for one time = $115

So budget remains =

So in $0.25 we get = 1 flyer

In $440 we get =  = 1760

Hence, we can buy 1760 flyers.

hope this helped

8 0
3 years ago
A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0
yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

P(x

Where, for example, P(0) and P(1) are calculated as:

P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098

Then, P(x, so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)

Where, for example, P(4) is calculated as:

P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001

Then, P(x\geq4 )=0.3504, so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0
4 years ago
How many solutins<br> 6x + 35 +9x =15x(x+4) -25
Sveta_85 [38]
Hi again............. I am glad to help you again.

6x + 35 + 9x = 15x(x+4) -25
6x + 35 + 9x = 15x² + 60x - 25
15x + 35 = 15x² + 60x - 25
Now, we gonna put all the common terms together
15x - 60x + 35 + 25 - 15x² =
-45x + 60 - 15x² = 0
Now, we need to factorize
-15(x-1)(x+4)=0
Set factors equal to 0
x -1 = 0 or x + 4 = 0
x = 1 or x = 4


I hope that's help ! 

5 0
3 years ago
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