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nalin [4]
3 years ago
7

F(x)=2(x)^2+5 sqrt (x+2) f(0)

Mathematics
1 answer:
krek1111 [17]3 years ago
7 0
F(0) = replace every x with 0

2 ({0})^{2}  + 5 \sqrt{0 + 2}  \\  = 2 + 5 \sqrt{2}
=9.0710678119
=9.07
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What does 3x2 + 9x − 30
Harlamova29_29 [7]

Answer:38

Step-by-step explanation:

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3 years ago
Compute the conditional probabilities from the two-way frequency table.
Natali5045456 [20]

Answer:

The table is show clearly in the figure attached.

P(boy if favourite activity is swimming) = 8/17 = 0.47

P(girl if favourite activity is sport) = 7/27 = 0.26

P(girl if favourite activity is reading) = 4/6 = 0.67

P(boy if favourite activity is sport) = 20/27 = 0.74

P(favourite activity is swimming if a girl) = 9/20 = 0.45

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6 0
3 years ago
What is sin 0 when sec 0 = 5? 0 |(1,0) sin e = 2[?] [] Rationalize the denominator if necessary.​
Komok [63]

Answer:

Sin ~0 =\frac{2\sqrt{5} }{5}

Step-by-step explanation:

sec0=\sqrt{5}

cos~0=\frac{1}{sec0} =\frac{1}{\sqrt{5} }

sin ~0=\sqrt{1-cos^2~0}

=\sqrt{1-1/5}=\sqrt{5-1/5}

Sin~0=\sqrt{4/5} =2/\sqrt{5}*\sqrt{5}/\sqrt{5}

Sin~0=2\sqrt{5}/2=2\sqrt{5}/5

<u>--------------------------------</u>

hope it helps...

have a great day!!!

7 0
3 years ago
During a sale, a store offered a 20% discount on a stereo system that originally sold for $320. After the sale, the discounted p
Phoenix [80]

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354 $ is correct

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3 0
3 years ago
In an experiment, a fair coin is tossed 13 times and the face that appears (H for head or T for tail) for each toss is recorded.
zalisa [80]

Answer:

1) 1 element

2) 13 elements

3) 22 elements

4) 40 elements

Step-by-step explanation:

1) Only one element will have no tails: the event that all the coins are heads.

2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.

3) There are {13 \choose 2} = 78 elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are {11 \choose 2} = 55 possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22

4) We can have:

  • A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
  • One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)

For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)

For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).

This gives us a total of 18+22 = 40 possibilities.

5 0
3 years ago
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