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3 years ago

The points R(3,−3), S(8,0), and T(4,1) are connected to form △RST.

Mathematics

2 answers:

denis-greek [22]3 years ago

7 0

Answer:

Isosceles.

Step-by-step explanation:

We are going to check it with the distance formula

d(R,T) = $\sqrt[2]{(3-4)^{2}+(-3-1)^{2}}$

= $\sqrt[2]{(-1)^{2}+(-4)^{2}}$

= $\sqrt[2]{1+16}$

= $\sqrt[2]{17}$ .

d(T,S) = $\sqrt[2]{(4-8)^{2}+(1-0)^{2}}$

= $\sqrt[2]{(-4)^{2}+(1)^{2}}$

= $\sqrt[2]{16+1}$

= $\sqrt[2]{17}$ .

d(S,R) = $\sqrt[2]{(8-3)^{2}+(0-(-3))^{2}}$

= $\sqrt[2]{(5)^{2}+(3)^{2}}$

= $\sqrt[2]{25+9}$

= $\sqrt[2]{34}$ .

Then, as the triangle has two sides with the same length, the triangle is isosceles.

Lady bird [3.3K]3 years ago

5 0

right isosceles

rs:34&rs=rt+st

rt:17

st:17

ab=[(x(a)-x(b))+(y(a)+y(b))]

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let X = number of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, p = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, n = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable X follows a Binomial distribution with parameter n = 55 and p = 0.20.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:

$E(X)=np$

$=55\times 0.20\\=11$

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$

$=0.1968$

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

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3 years ago

Matthew haa 18 sets of baseball cards each st has 12 cards about how many baseball card does matthew have

cluponka [151]

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Heron Island is a coral cay in the Great Barrier Reef. Around Heron Island (and elsewhere in Australia) there are 2 color morphs

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3 years ago

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kow [346]

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Step-by-step explanation:

s = 7

k = 4

x = 6

7 - 4 + 6

3 + 6

8 0

3 years ago