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2 years ago

PLS HELP DUE SOON: (basically it's a retest) will give brainliest <3

Mathematics

2 answers:

Varvara68 [4.7K]2 years ago

4 0

If I’m correct I believe it’s 4.

denis-greek [22]2 years ago

3 0

The answer is the fourth one

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2 Classify the triangle by its sides. The diagram is not to scale. A straight B equilateral C scalene D isosceles (sides are 8,6

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Answer:

D) Isosceles

Step-by-step explanation:

An isosceles triangle is the triangle that has 2 sides that are the same.

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2 years ago

A rational function can have infinitely many x-values at which it is not continuous. True or False. If false, explain

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Each inch on a map of Roswell, New Mexico represents $0.51\text{ mile}$. If two hills in Roswell are $7$ inches apart on the map

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Answer:

$3.57$

Step-by-step explanation:

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$\frac{1}{0.51}\frac{inch}{miles}$

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2 years ago

Chelsea is graphing the function f(x) = 20(One-fourth)x. She begins by plotting the initial value. Which graph represents her fi

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2 years ago

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A random draw is being designed for 210 participants. A single winner is to be chosen, and all the participants must have an equ

Over [174]

Answer: The correct number of balls is (b) 4.

Step-by-step explanation: Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.

We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.

Let 'r' represents the number of balls to be picked up.

Since we are choosing from 10 balls, so we must have

$^{10}C_r=210.$

The value of 'r' can be any one of 0, 1, 2, . . , 10.

Now,

if r = 1, then

$^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10$

If r = 2, then

$^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45$

If r = 3, then

$^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120$

If r = 4, then

$^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.$

Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.

Thus, (b) is the correct option.

4 0

2 years ago

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