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Ierofanga [76]
3 years ago
5

During which phase of the moon do we see the entire lighted side of the moon? first quarter new moon full moon waning gibbous

Mathematics
1 answer:
Fudgin [204]3 years ago
4 0

Answer: Full Moon

Step-by-step explanation:

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Each interior angle of a regular polygon has a measure of 40.how many sides does the polygon have?​
zysi [14]

<u>Given</u><u> </u><u>Information</u><u> </u><u>:</u><u>-</u>

⠀

  • Each exterior angle of a regular polygon has a measure of 40°

⠀

<u>To</u><u> </u><u>Find</u><u> </u><u>:</u><u>-</u>

⠀

  • The number of sides of the polygon

⠀

<u>Formula</u><u> </u><u>Used</u><u> </u><u>:</u><u>-</u>

⠀

\qquad \star \:  \underline{ \boxed{ \green{  \sf No.~of~sides = \dfrac{360^ \circ}{exterior ~angle}}}} \:  \star

⠀

<u>Solution</u><u> </u><u>:</u><u>-</u>

⠀

Using the formula,

⠀

\sf \dashrightarrow No. ~of~sides= \dfrac{360}{40}  \\  \\  \\ \sf \dashrightarrow No. ~of~sides= \frac{ \cancel{360}}{ \cancel{40} } \\  \\  \\  \sf \dashrightarrow No. ~of~sides= \underline{ \boxed{ \blue { \frak{9}}}} \:  \star

⠀

Thus, the polygon is a nonagon, and hence has 9 sides.

⠀

\underline{\rule{227pt}{2pt}} \\  \\

5 0
2 years ago
For an outdoor event organizers estimate that 60000 people will attend if it is not raining. If it is raining, event organizers
s2008m [1.1K]

For an outdoor event organizers, the expected number of people who will attend is 32700

<u>SOLUTION:</u>

Given, For an outdoor event organizers estimate that 60000 people will attend if it is not raining.  

And If it is raining, event organizers estimate that 21000 people will attend.  

On the day of the show, meteorologists predicted a 70% chance of rain.

We have to determine the expected number of people who will attend.

Now, number of people = 70% of rain case + 30% of other case.

Number of people = 0.7 x rain case + 0.3 x other case

Number of people = 0.7 x 21000 + 0.3 x 60000

= 2100 x 7 + 6000 x 3 = 32700.

Hence, the expected count is 32700

3 0
2 years ago
A die with 8 sides is rolled.what is the probability of rolling a number less than 7
larisa86 [58]
We want that number to be 1 2 3 ....6

the probability is = 6/8 we simplify by 2=3/4
3 0
3 years ago
Find the Inverse of 3x^3-1=y
Anika [276]
y = 3x^{3} - 1
x = 3y^{3} - 1
3y^{3} = x + 1
y^{3} = \frac{x + 1}{3}

y = \sqrt[3]{\frac{x + 1}{3}}
5 0
3 years ago
A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f
Mrrafil [7]

Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

3) The contestant gets the question 2 and gets the question 1 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

3 0
2 years ago
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