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3 years ago

HELPPPP PLEASE!!!! Don't understand

Mathematics

2 answers:

iragen [17]3 years ago

7 0

Answer: 20% is your answer

Step-by-step explanation:

180-150=30

30 divided by 150 = .2 which is 20%

laiz [17]3 years ago

4 0

Answer:

The answer is B. 20%

It helps to use properties of multiplication and division.

Also, you can think of 20% as $\frac{20}{100}$

Hope this helps

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What are all the possible rectangles with whole-number side

11111nata11111 [884]

Answer:

So (L,W) possibilities are:

(1,4),(4,1),(2,3),(3,2)

That makes 4 possibilities.

Step-by-step explanation:

The perimeter of a rectangle is P=2L+2W where L is the length and W is the width.

We have that P=10, so 10=2L+2W.

10=2L+2W

10=2(L+W) By factoring using the distributive property.

2(5)=2(L+W) I factored 10 as 2(5).

If 2(5)=2(L+W), then 5=L+W.

Whole numbers are {0,1,2,3,4,5,6,7,8,9,10,...}. They are your counting numbers and 0.

I think they want natural numbers {1,2,3,4,...}. This is also just called the counting numbers. The reason I think they want this because if one of the dimensions is 0, we won't actually have a rectangle.

So now looking for numbers from this set that satisfy: L+W=5.

L+W=5

1+4=5

4+1=5

2+3=5

3+2=5

So (L,W) possibilities are:

(1,4),(4,1),(2,3),(3,2)

That makes 4 possibilities.

5 0

3 years ago

Triangle XYZ Is similar to triangle PGR

marishachu [46]

Answer:

I think 24 will be ur ans_______

7 0

3 years ago

You have a biased coin for which p(h)=pp(h)=p. you toss the coin 2020 times. what is the probability that you observe 88 heads a

Nat2105 [25]

The question describes a binomial probability with p(h) = p, then p(t) = 1 - p and number of trials (n) = 20

The probability of a binomial distribution is given by

$P(x)=\, ^nC_xp^x(1-p)^{n-x}$

Part A:

The probability of observing 8 heads and 12 tails is given by:

$P(8)=\, ^{20}C_8p^8(1-p)^{20-8}=\, ^{20}C_8p^8(1-p)^{12}$

Part B:

You observe more than 8 heads and more than 8 tails, when you observe 9 heads and 11 tails, 10 heads and 10 tails, and 11 heads and 9 tails.

Therefore, the probability of observing more than 8 heads and more than 8 tails is given by:

$P(9)+P(10)+P(11) \\ \\ =\, ^{20}C_9p^9(1-p)^{20-9}+\, ^{20}C_{10}p^{10}(1-p)^{20-10}+\, ^{20}C_{11}p^{11}(1-p)^{20-11} \\ \\ =\, ^{20}C_9p^9(1-p)^{11}+\, ^{20}C_{10}p^{10}(1-p)^{10}+\, ^{20}C_{11}p^{11}(1-p)^{9}$