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Semmy [17]
3 years ago
15

Problem #5

Mathematics
1 answer:
goblinko [34]3 years ago
8 0
25% P x V1 = V2 P= 1.5/30=0.05x 100
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Wendy throws a dart at this square-shaped target:
Ostrovityanka [42]

Answer:

Hello! I hope I am correct! :)

Step-by-step explanation:

Let’s first calculate the black circle & the white space.

Since there is more white space than the black circle, we already know that the white space will have more probability.

These are the steps you need to do, in order to so love this problem:

1. Find the area of both, the white space and the black circle.

2. Divide both of them by the area of the square.

3. Do all these steps to find the probability/hitting.

Part A:

Black circle: ( π *1^2) / ( π 5^2) = 1/25 = .04 or 3.14% chance.

So we can tell it’s close to zero.

Part B: 1 - .04 = 0.96 or 97%

This is close to one.

(Remember I am not saying the exact, it’s a estimate)

Brainliest would be appreciated!

Hope this helps! :)

By; BrainlyAnime

8 0
3 years ago
In a box plot, if there is an outlier shown in the picture below, what is considered the min? The outlier or 59?
scZoUnD [109]

Step-by-step explanation:

The outlier is excluded, so the minimum is 59.

5 0
3 years ago
At Jack’s birthday party, 23 of the kids had balloons. 14 of the balloons were red. What fraction of the kids at the party had r
Nonamiya [84]
14/23. You can’t simplify it anymore. It is also equivalent to about 61% (60.86% to be exact)
7 0
3 years ago
Evaluate the expression where x=5 and y =2
solong [7]

Answer:

6

Step-by-step explanation:

8 0
3 years ago
WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"
Feliz [49]

Recall Euler's theorem: if \gcd(a,n) = 1, then

a^{\phi(n)} \equiv 1 \pmod n

where \phi is Euler's totient function.

We have \gcd(9,32) = 1 - in fact, \gcd(9,32^k)=1 for any k\in\Bbb N since 9=3^2 and 32=2^5 share no common divisors - as well as \phi(9) = 6.

Now,

37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} =  32\cdot\left(32^{(\cdots)}\right)^6

where the c_i are positive integer coefficients from the binomial expansion. By Euler's theorem,

\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9

so that

32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9

7 0
1 year ago
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