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RoseWind [281]
4 years ago
8

PLS HELP find the unknown side of the triangle below (round to the nearest tenth) ​

Mathematics
1 answer:
Nonamiya [84]4 years ago
7 0
22.9 is the closest but not exact. It was probably rounded off.

I used the formula: leg^2+leg^2=hypotenuse^2

(You can use that formula for any right triangle.)
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Please help!! Tysm!!!!!!!!!!
kolezko [41]
It’s the first one.

the one with 9 squares with 2 filled in
7 0
3 years ago
The equation we are given (-at² + bt + c) is a parabola. How do we know this?
Scrat [10]

It's a polynomial of degree 2. Every polynomial of degree 2 is a parabola

4 0
3 years ago
Round 392 to the nearest ten
Bezzdna [24]
Nearest 10 is according to the last digit.
if the digit is >5, it +1 to the tenth digit. If it's <5, it will be reduced to 0.

for your case, 392, 2 < 5, therefore rounded up to nearest tenth is 390.
5 0
3 years ago
Read 2 more answers
Let X be the temperature in at which a certain chemical reaction takes place, and let Y be the temperature in (so Y = 1.8X + 32)
Black_prince [1.1K]

Answer:

See explanation

Step-by-step explanation:

Solution:-

The random variable, Y be the temperature of chemical reaction in degree fahrenheit be a linear expression of a random variable X : The  temperature in at which a certain chemical reaction takes place.

                             Y = 1.8*X + 32

- The median of the random variate "X" is given to be equal to "η". We can mathematically express it as:

                             P ( X ≤ η ) = 0.5

- Then the median of "Y" distribution can be expressed with the help of the relation given:

                             P ( Y ≤ 1.8*η + 32 )

- The left hand side of the inequality can be replaced by the linear relation:

                             P ( 1.8*X + 32 ≤ 1.8*η + 32 )

                             P ( 1.8*X ≤ 1.8*η )   ..... Cancel "1.8" on both sides.

                            P ( X ≤ η ) = 0.5 ...... Proven

Hence,

- Through conjecture we proved that: (1.8*η + 32) has to be the median of distribution "Y".

b)

- Recall that the definition of proportion (p) of distribution that lie within the 90th percentile. It can be mathematically expressed as the probability of random variate "X" at 90th percentile :

                             P ( X ≤ p_.9 ) = 0.9 ..... 90th percentile

- Now use the conjecture given as a linear expression random variate "Y",

          P ( Y ≤ 1.8*p_0.9 + 32 ) = P ( 1.8*X + 32 ≤ 1.8*p_0.9 + 32 )

                                                 = P ( 1.8*X ≤ 1.8*p_0.9 )

                                                 = P ( X  ≤ p_0.9 )

                                                 = 0.9

- So from conjecture we saw that the 90th percentile of "X" distribution is also the 90th percentile of "Y" distribution.

c)

- The more general relation between two random variate "Y" and "X" is given:

                            Y = aX + b

Where, a : is either a positive or negative constant.

- Denote, (np) as the 100th percentile of the X distribution, so the corresponding 100th percentile of the Y distribution would be : (a*np + b).

- When a is positive,

                   P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

                                                 = P ( a*X ≤ a*p_% )

                                                 = P ( X  ≤ p_% )

                                                 = np_%        

- When a is negative,

                   P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

                                                 = P ( a*X ≤ a*p_% )

                                                 = P ( X  ≥ p_% )

                                                 = 1 - np_%        

                                                           

4 0
3 years ago
You have also set up a card game in which a player picks a card from a standard deck of 52 cards. The player wins if these two e
Alenkinab [10]

First, let's count:

there are 26 possible outcomes for E1 (black card)

there are 4x9 = 36 possible outcomes for E2, to pick a numbered card (any color)

there are 2x9 =18 possible outcomes for E1 (black) AND E2 (numbered, spade + clower)

the probability of E1 AND E2 is the ratio of the count of possible outcomes for E1 + E2 and the count of all possible outcomes (52 choices to pick a card from the deck):

P(E1 and E2) = 18/52 (34.6%)

And as asked:

P(E1) = 26/52 = 1/2 (50%)

P(E2) = 36/52 = 9/13 (69.2%)

5 0
3 years ago
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