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dexar [7]
3 years ago
11

I'm learning about sin, cos, and tan but I don't know how to use my calculator. My problem right now is -1.8=cos x and I need to

solve for x but I don’t know what to do in my calculator.
Mathematics
2 answers:
kvasek [131]3 years ago
8 0

Answer:

can you type out the entire equation? i'll do my best to solve it. don't screen shot the question tho because for some reason my laptop won't let me see them.

Step-by-step explanation:

Reil [10]3 years ago
7 0

Answer:

there is no solution for x

Step-by-step explanation:

BTW you can use math.way

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Cluster Sampling

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In the rectangular prism shown below, which lines are parallel?
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JN and LP...............

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Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t
LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6)  =\dfrac{20}{36}

The probability distribution of this event is given as follows.

\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|

First, we determine the expected Value of this event.

Expected Value

=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning  =\dfrac{20}{36}

If the game is played 100 times

Number of times expected to win

=\dfrac{20}{36} \times 100\\=56$ times

Therefore, number of times expected to loose

= 100-56

=44 times

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Answer: 90 projects in all 1/5 were photographs 1/5 of 90=90/5=18

Step-by-step explanation: done :)

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