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Elan Coil [88]
3 years ago
15

Plz help illl give brainlist to you

Mathematics
2 answers:
harina [27]3 years ago
7 0
33 girls and 57 boys
so total = 33 + 57 = 90 players

20% left handed = 90 x 0.2 = 18

answer
18 left-handed players
yulyashka [42]3 years ago
3 0
Add all the players first 33 + 57 = 90
Use the 20% and then multiply it by 90
20% x 90 = 18
18 Kids should be your answer
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The plot below shows the graphs of three functions, f, g, and h, all of which determine the population of three different cities
monitta

So, here we have an exponential function.

Remember that an exponential function has the form:

y=a(b)^x=a(1\pm\frac{r}{100})^x

Where a represents an initial amount, and r is the rate of this amount to change. (Increase, or decrease).

So, given that the population of City A in 2000 was 40 thousand people and the population increased by 13% each year, we can say that

\begin{gathered} a=40 \\ b=1+\frac{13}{100}=1.13 \end{gathered}

So,

f(x)=40(1.13)^x

For city B:

\begin{gathered} a=40 \\ b=1+\frac{16}{100}=1.16 \\ g(x)=40(1.16)^x \end{gathered}

But something different happens with city C. This is not an exponential function, this is a linear function.

So,

h(x)=40+10x

7 0
1 year ago
How many year apart is 384 BC from 1952 AD
julsineya [31]
BC and BCE are Before Common Era
So what you want to do is add 1952 and 384
1952 + 384 = 2336
So the amount of years between 384 BC and 1952 AD is about 2,336 years
Hope this helps
6 0
3 years ago
Read 2 more answers
Kira decorates the exterior faces of a gift box in the shape of a cube .What is the surface area in square inches of the gift bo
Gekata [30.6K]

Answer:

The surface area of the gift box that Kira decorates is 253.5\ in^{2}

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

A cube has six square faces

The surface area of a cube is equal to the area of the six squares faces

SA=6b^{2}

where

b is the length side of a cube

In this problem we have

b=6.5\ in

substitute

SA=6(6.5)^{2}

SA=253.5\ in^{2}

7 0
3 years ago
How to I solve problem 1?
Verdich [7]
#1

The uniforms are numbered 0, 1, 2, ..., 99. That's 100 numbers. Half of them are odd and half of them are even. So the probability that any one of the uniforms is odd is 1/2 just like the probability that any one uniform is even is 1/2.

(a) The numbers on the uniforms are independent of one another. That is, the number of her cross-country uniform does not in any way determine the number on her basketball uniform and vice versa. This means that we can find the probability that each is odd and multiply these together using what is called the counting principle. The probability that all are odd is:
(1/2)(1/2)(1/2)=1/8

(b) This is done the same way we did part (a). Since the probability of any one uniform being odd is the same as it being even (1/2), the answer here is the same: (1/2)(1/2)(1/2)=1/8

(c) This problem differs from that in (a) and (b). There is only one way for all three uniforms to be odd numbers: (odd, odd, odd) or all even (even, even, even). However, there are multiple ways for the uniforms to be two odd and one even. If the uniforms are listed in order: cross-country, basketball, softball we can get exactly one even in any of three ways:
even, odd, odd
odd, even, odd
odd, odd, even
The probability for any one of these possibilities is (1/2)(1/2)(1/2)=1/8 but since there are three way the probability that we get even exactly once is equal to (3)(1/8) = 3/8
7 0
3 years ago
What is the value of k in the equation k\3 -5=34
Alex
If wrong i apologize but for that equation i got the answer K=117
4 0
3 years ago
Read 2 more answers
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