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3 years ago

$229.95 with 25% off (SHOW WORK)

Mathematics

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

$172.46

Step-by-step explanation:

Discount = 229.95 × 25/100

Discount = 229.95 x 0.25

= $57.49

Final Price = 229.95 - 57.4875

= $172.46

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20+30-10+40=80<br>true or false

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Which describes a non-uniform probability model? A) heads or tails from a coin flip B) the sum result of rolling two dice C) sel

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Answer:

Option B.

Step-by-step explanation:

The non-uniform probability is when the event is not same in the occurring of event.

Option A is uniform because there will be equal chance of getting head or tail.

Option B is non-uniform because sum can be any number if we get(3,4) it will be 7 and if we have (2,3) then it will be 5.

Option C is uniform because we have equal number of red and green balls so, selecting a ball from both will be same.

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3 years ago

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3 years ago

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2 years ago

Anyone knows how to do questions 7 and 8? 15 pts!!

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7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: $2x^2-4x+5$ . With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

$E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3$

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

$2x^2+5=4x\\\\ 2x^2-4x+5=0$

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

$E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)$

Multiplying by (2-1) in the both sides:

$(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ...$ Repeating the process, we obtain: $...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}$

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