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Kazeer [188]
3 years ago
12

At a basketball game, a team made 52 successful shots. They were a combination of 1- and 2-point

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

17 for 1 point shots and 35 for two point shots.

Step-by-step explanation:

let the number of 1 point shots be x

let the number of 2 point shots be y

 

x + y = 52          (1)

x + 2y = 87        (2)

 

subtracting (1) from (2)

 

y = 35

 

from (1) x = 52 - y = 52 - 35 = 17

 

Hence number of 1 pointers = 17

 

number of 2 point shots = 35

 

checking; 17 + 35 =52 ; and 17+ 2*35 = 87

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Y=-5x + 22<br> -2x - 8y = 14
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Step-by-step explanation:

Put the value y.

-2x-8(-5x+22)=14

-2x+40x-176= 14

38x-176 =14

38x = 190

x=5

Now put the value of X to find Y

Y= -5(5)+22= -25+22 = -3

=>X=. 5

=>Y= -3

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Prove that the diagonals of a parallelogram bisect each other.<br> The midpoint of AC is
iren2701 [21]

Answer:

Theorem: The diagonals of a parallelogram bisect each other. Proof: Given ABCD, let the diagonals AC and BD intersect at E, we must prove that AE ∼ = CE and BE ∼ = DE. The converse is also true: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Step-by-step explanation:

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3 years ago
Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the
Likurg_2 [28]

Answer:

(a)  moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, P(X)=\frac{1}{6}

(a) Moment generating function can be written as M_{x}(t).

M_x(t)=\sum_{x=1}^{6} P(X=x)

M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}

M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) Now, find E(X) \text { and } E\((X^{2}) using moment generating function

M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)

M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)  

\Rightarrow E(X)=\frac{21}{6}

M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)

M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)

\Rightarrow E\left(X^{2}\right)=\frac{91}{6}  

Hence, (a) moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right).

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

6 0
4 years ago
Which sequences of transformations would result in similar figures that are enlargements or reductions?
Ugo [173]

Answer:

Dilations

Step-by-step explanation:

8 0
3 years ago
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