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3 years ago

Is the relation shown in the arrow diagram a function?

Mathematics

1 answer:

Irina-Kira [14]3 years ago

4 0

Answer:

Yes. Every unique input has a unique output.

Step-by-step explanation:

Since every input has a different output, therefore it's a function. It's not a function when the input have more than 1 output. When graphing, make sure you take the vertical line test to see whether or not the graph is a function.

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5/9 × 8/11 what is the answer

nikitadnepr [17]

Answer:

$\frac{5}{9} \times \frac{8}{11} = \frac{5 \times 8}{9 \times 11} = \frac{40}{99} = 0.4040$

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3 years ago

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54 students fit on a bus. There are 125

vfiekz [6]

Answer: 3 buses

Step-by-step explanation: If 54 students fit on a bus, and 125 students are in 8th grade,

You should divide.

125 divided by 54 is 2.31...

we can't have 2.31... buses, so we say 3 buses.

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2 years ago

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Using approciate properties find -4/7*3/5+4/7*-1/5+2/5*-4/5

slega [8]

Answer: 0.548

Step-by-step explanation:

First you need to remember the formula for PEMDAS(Parenthesis,exponents,multiplication,divison,addition,subtractio)

you can do either multiplication or divison whichever comes first, same with addition+subtraction.

the underlined means to simplify

1.-0.57x3/5+4/7x-1/5+2/5x-4/5

2.-1.71/5+4/7x-1/5+2/5x-4/5

3.-0.342+4/7x-1/5+2/5x-4/5

4.-0.342+0.57x-1/5+2/5x-4/5

5.-0.342+0.57x-0.2+2/5x-4/5

6.-0.342+0.114+2/5x-4/5

7.-0.342+0.114+0.4x-4/5

8.-0.342+0.114+-1.6/5

9.-0.342+0.114+-0.32

10.-0.228+-0.32=-0.548

i did this all in one go so......

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3 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

3 years ago

PLEASE HELP DUE SOON MORE POINTS IF CORRECT!!

SVETLANKA909090 [29]

Answer:

y=4x+35

Step-by-step explanation:

5 0

3 years ago

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